Geometric series are the one large family whose sums are known exactly, but they rarely arrive written as . How do you see one through its disguise, read off its first term and ratio, and use it to judge the series around it?

Calculus 2

The geometric series test

Geometric series are the one large family whose sums are known exactly, but they rarely arrive written as \sum a r^n. How do you see one through its disguise, read off its first term and ratio, and use it to judge the series around it?

You need

Infinite series and partial sumsCalculus 2

Unlocks

Representing functions as power seriesCalculus 2

Where you are 8 in Sequences and series in calculus

Calculus8

Sequences and series in calculus8

Convergence of sequences Infinite series and partial sums

The geometric series test

The nth term test for divergence Convergence tests6 ▾ Show all 8 (3 more)

Why this exists

Oyunaa strikes the A below middle C on a piano, a string that vibrates 220 times a second. The hammer leaves 50 millijoules of energy in the string, and in every cycle the string loses 0.8\% of the energy it holds at the start of that cycle, to the soundboard and the air. The energy given off in cycle n, in millijoules, is

50(0.008)(0.992)^{n-1} = 0.4(0.992)^{n-1}

and the total given off is the series \sum_{n=1}^{\infty} 0.4(0.992)^{n-1}. Each term is the one before it times 0.992: the series is geometric.

Most series have no formula for their partial sums, and Infinite series and partial sums could find one only by telescoping. Geometric series are the large family where the partial sums are known exactly, so they are the first thing to look for in a new series. They seldom arrive in the form \sum a r^n, though. They come as \displaystyle \frac{5^{n+1}}{7^{n-1}}, as \displaystyle \frac{(-2)^n}{3^{2n+1}}, starting at n = 2, or added to another geometric series. This lesson is about seeing them, reading off the two numbers that decide them, and using them as the yardstick that the later convergence tests measure other series against.

Definition

In plain words: a series is geometric when each term is the previous term times one fixed number.

A series \sum_{n=m}^{\infty} b_n with every b_n \ne 0 is geometric with ratio r if, for every n \ge m,

\frac{b_{n+1}}{b_n} = r

Its first term is a = b_m, the term at the starting index, and then b_n = a r^{n-m} for every n \ge m.

The geometric series test. A geometric series with first term a \ne 0 and ratio r converges if \lvert r \rvert < 1, with sum

\sum_{n=m}^{\infty} a r^{n-m} = \frac{a}{1-r}

and diverges if \lvert r \rvert \ge 1. In words: the sum is the first term over one minus the ratio, whatever the starting index.

Intuition

The partial sums of a geometric series are S_N = a\dfrac{1 - r^{N}}{1 - r} when the series has N terms so far (r \ne 1), and the only part that moves with N is r^N. Everything turns on whether powers of r die out. In Figure 1 they do: 0.992^n shrinks slowly, and Oyunaa’s running total of energy given off climbs toward 50 millijoules, which is all the energy the hammer put in. That total had to be 50, and the formula agrees: \displaystyle \frac{0.4}{1 - 0.992} = 50.

Figure 1The energy Oyunaa’s string has given off after n cycles, S_n = 50\left(1 - 0.992^n\right) millijoules. The partial sums of a geometric series with ratio 0.992 rise toward 50 and never pass it.

Recognition rests on one fact. A term of the form C d^{\,n}, a constant times a fixed number to the power n, is geometric with ratio d, because \displaystyle \frac{C d^{\,n+1}}{C d^{\,n}} = d. Exponents of the form kn + j fit this mold, since q^{kn+j} = q^{j}\left(q^{k}\right)^n, and so do products and quotients of such powers. So the work on a disguised series is algebra: collect every power of n into a single d^{\,n}, and what is left over is the constant.

Results

Recognition. If b_n = C d^{\,n} for constants C \ne 0 and d \ne 0, then \sum_{n=m}^{\infty} b_n is geometric with ratio d and first term C d^{\,m}. Conversely every geometric series has this form, with C = a r^{-m} and d = r.

Any starting index. \sum_{n=m}^{\infty} b_n converges to \dfrac{b_m}{1 - r} when \lvert r \rvert < 1. The first term is b_m, the term at the starting index, not the constant C.

Two geometric series. Let A, B, p, q be nonzero with p \ne q. Then \sum_{n=m}^{\infty} \left(A p^n + B q^n\right) converges exactly when \lvert p \rvert < 1 and \lvert q \rvert < 1, and then its sum is the sum of the two separate sums.

Every geometric series telescopes. For d \ne 1, put c_n = \dfrac{C d^{\,n}}{1 - d}. Then

C d^{\,n} = c_n - c_{n+1}

A telescoping series with c_n \to 0 need not be geometric, so telescoping is the wider family, and geometric series are the members whose c_n is itself geometric.

Why it is true

The test. This is proved in Infinite geometric series and only recalled here. Subtracting r S_N from S_N = a + ar + \cdots + ar^{N-1} leaves (1 - r)S_N = a(1 - r^N). When \lvert r \rvert < 1, r^N \to 0 and \displaystyle S_N \to \frac{a}{1-r}; when \lvert r \rvert \ge 1, the terms a r^n do not tend to 0 and the The nth term test for divergence gives divergence.

Recognition and the starting index. If b_{n+1} = r b_n for every n \ge m, then b_{m+1} = r b_m, b_{m+2} = r^2 b_m, and by induction b_n = b_m r^{n-m}. Renaming k = n - m turns \sum_{n=m}^{\infty} b_m r^{n-m} into \sum_{k=0}^{\infty} b_m r^{k}, the same series counted from 0, so the first term is b_m.

Two geometric series. If \lvert p \rvert < 1 and \lvert q \rvert < 1, both series converge and the sum law for series adds them. If \lvert p \rvert < 1 \le \lvert q \rvert and the combined series converged, then \sum B q^n would be the difference of two convergent series and would converge, which it does not. If \lvert p \rvert \ge 1 and \lvert q \rvert \ge 1, say \lvert p \rvert \ge \lvert q \rvert, there are two cases. When \lvert p \rvert > \lvert q \rvert, the term is p^n\left(A + B (q/p)^n\right); the bracket tends to A \ne 0 and \lvert p^n \rvert \ge 1, so the terms do not tend to 0. When \lvert p \rvert = \lvert q \rvert and p \ne q, then q = -p and the term is p^n\left(A + B(-1)^n\right). The numbers A + B and A - B cannot both be 0 (that would force A = 0), so infinitely many terms have size at least the nonzero one of \lvert A + B \rvert, \lvert A - B \rvert. Either way the series diverges by the The nth term test for divergence.

Telescoping. c_n - c_{n+1} = \dfrac{C d^{\,n} - C d^{\,n+1}}{1 - d} = \dfrac{C d^{\,n}(1 - d)}{1 - d} = C d^{\,n}.

Worked examples

Example 1

Oyunaa’s piano string

The energy Oyunaa’s string gives off in cycle n is 0.4(0.992)^{n-1} millijoules. Find the total over all cycles, and the energy given off from cycle 100 on.

Solution

The ratio of consecutive terms is 0.992, so the series is geometric, and \lvert 0.992 \rvert < 1, so it converges. Starting at n = 1 the first term is 0.4(0.992)^0 = 0.4, and

\sum_{n=1}^{\infty} 0.4(0.992)^{n-1} = \frac{0.4}{1 - 0.992} = \frac{0.4}{0.008} = 50

millijoules: in the long run the string gives off all the energy the hammer put in.

From cycle 100 on, the series \sum_{n=100}^{\infty} 0.4(0.992)^{n-1} has the same ratio and a new first term, the term at n = 100, which is 0.4(0.992)^{99}. So

\begin{aligned} \sum_{n=100}^{\infty} 0.4(0.992)^{n-1} &= \frac{0.4(0.992)^{99}}{0.008} \\ &= 50(0.992)^{99} \approx 22.57 \end{aligned}

millijoules. That is the energy still in the string after 99 cycles, as it must be, since all of it is eventually given off. At 220 cycles a second this is less than half a second after the strike.

Example 2

Which of these are geometric?

Decide which series are geometric, and settle each one.

(a) \sum_{n=1}^{\infty} e^{3-2n} (b) \sum_{n=0}^{\infty} \cos(n\pi)\,(0.9)^n (c) \sum_{n=1}^{\infty} \dfrac{4^n}{n} (d) \sum_{n=1}^{\infty} \left(3^n + 2\right)

Solution

(a) e^{3-2n} = e^3\left(e^{-2}\right)^n, a constant times a power, so the ratio is e^{-2} \approx 0.135 and the series converges. Its first term, at n = 1, is e^{1} = e, and the sum is \dfrac{e}{1 - e^{-2}} = \dfrac{e^3}{e^2 - 1}, multiplying top and bottom by e^2.

(b) \cos(n\pi) = (-1)^n for every integer n, so the term is (-1)^n(0.9)^n = (-0.9)^n. The ratio is -0.9, the first term at n = 0 is 1, and the sum is \dfrac{1}{1 + 0.9} = \dfrac{10}{19}.

(c) The ratio of consecutive terms is \dfrac{4^{n+1}/(n+1)}{4^n/n} = \dfrac{4n}{n+1}, which depends on n, so the series is not geometric. Its terms grow without bound, so it diverges by the The nth term test for divergence.

(d) The ratio \dfrac{3^{n+1} + 2}{3^n + 2} is \displaystyle \frac{11}{5} at n = 1 and \displaystyle \frac{29}{11} at n = 2, so the series is not geometric. It is the sum of the geometric series \sum 3^n and \sum 2 \cdot 1^n, ratios 3 and 1, and it diverges because its terms grow without bound.

Example 3

Exponents that shift: a start at n = 2

Decide whether \displaystyle \displaystyle\sum_{n=2}^{\infty} \frac{5^{n+1}}{7^{n-1}} converges, and if it does, find its sum.

Solution

Split each power into a constant and an nth power: 5^{n+1} = 5 \cdot 5^n and \displaystyle 7^{n-1} = \frac{7^n}{7}. So

\begin{aligned} \frac{5^{n+1}}{7^{n-1}} &= \frac{5 \cdot 7 \cdot 5^n}{7^n} \\ &= 35\left(\frac{5}{7}\right)^n \end{aligned}

The ratio is \displaystyle \frac{5}{7}, which lies in (-1, 1), so the series converges. The first term is the term at n = 2, not 35:

b_2 = \frac{5^{3}}{7^{1}} = \frac{125}{7}

and the sum is

\frac{125/7}{1 - 5/7} = \frac{125}{7} \cdot \frac{7}{2} = \frac{125}{2}

With the roles of 5 and 7 exchanged, \displaystyle \sum_{n=2}^{\infty} \frac{7^{n+1}}{5^{n-1}} has ratio \displaystyle \frac{7}{5} > 1 and diverges. The shifts in the exponents change only the constant, never the verdict.

Example 4

A negative base and an exponent of 2n

Find \displaystyle \displaystyle\sum_{n=0}^{\infty} \frac{(-2)^n}{3^{2n+1}}.

Solution

The 2n in the exponent hides the ratio: 3^{2n+1} = 3 \cdot \left(3^2\right)^n = 3 \cdot 9^n. So

\frac{(-2)^n}{3^{2n+1}} = \frac{1}{3}\left(-\frac{2}{9}\right)^n

The ratio is \displaystyle -\frac{2}{9}, not \displaystyle -\frac{2}{3}, and \displaystyle \left\lvert -\frac{2}{9} \right\rvert < 1. The first term, at n = 0, is \displaystyle \frac{1}{3}, so the sum is

\frac{1/3}{1 + 2/9} = \frac{1}{3} \cdot \frac{9}{11} = \frac{3}{11}

Because the ratio is negative, the partial sums \displaystyle \frac{1}{3}, \frac{7}{27}, \frac{67}{243}, \ldots fall on alternate sides of \displaystyle \frac{3}{11} \approx 0.2727, closing in from above and below.

Example 5

Khulan’s sum of two geometric series

Khulan meets \displaystyle \displaystyle\sum_{n=1}^{\infty} \frac{4 \cdot 3^n - 2^{n+1}}{6^n}. Is it geometric? Find its sum. Then decide \displaystyle \displaystyle\sum_{n=0}^{\infty} \frac{3^n + 5^n}{4^n}.

Solution

The first three terms are \displaystyle \frac{4}{3}, \displaystyle \frac{7}{9} and \displaystyle \frac{23}{54}, with ratios \displaystyle \frac{7}{12} and \displaystyle \frac{23}{42}, so the series is not geometric. Split the fraction instead:

\begin{aligned} \frac{4 \cdot 3^n - 2 \cdot 2^n}{6^n} &= \frac{4 \cdot 3^n}{6^n} - \frac{2 \cdot 2^n}{6^n} \\ &= 4\left(\frac{1}{2}\right)^n - 2\left(\frac{1}{3}\right)^n \end{aligned}

Both pieces are geometric series with ratio less than 1 in size, so both converge, and the sum law for convergent series allows adding them separately. From n = 1, \displaystyle \sum \left(\frac{1}{2}\right)^n = \frac{1/2}{1/2} = 1 and \displaystyle \sum \left(\frac{1}{3}\right)^n = \frac{1/3}{2/3} = \frac{1}{2}, so the sum is

4(1) - 2\left(\frac{1}{2}\right) = 3

The second series splits as \displaystyle \left(\frac{3}{4}\right)^n + \left(\frac{5}{4}\right)^n. The first piece converges and the second diverges, so by the result on two geometric series the whole series diverges. Here the terms themselves grow without bound, since \displaystyle \left(\frac{5}{4}\right)^n \to \infty.

Example 6

Telescoping, geometric, or both

Find \displaystyle\sum_{n=0}^{\infty} \left(3^{-n} - 3^{-n-1}\right) in two ways. Then find \displaystyle \displaystyle\sum_{n=1}^{\infty} \frac{n-1}{2^{n+1}} and show that it is not geometric.

Solution

As a telescoping series. With c_n = 3^{-n} the term is c_n - c_{n+1}, so the partial sum through n = N collapses:

S_N = c_0 - c_{N+1} = 1 - 3^{-N-1} \to 1

As a geometric series. Factor out 3^{-n}: \displaystyle 3^{-n} - 3^{-n-1} = 3^{-n}\left(1 - \frac{1}{3}\right) = \frac{2}{3}\left(\frac{1}{3}\right)^n. The ratio is \displaystyle \frac{1}{3}, the first term at n = 0 is \displaystyle \frac{2}{3}, and the sum is \displaystyle \frac{2/3}{1 - 1/3} = 1. The two methods agree, as they must.

For the second series the first terms are \displaystyle 0, \frac{1}{8}, \frac{1}{8}, \frac{3}{32}. The ratio from the second term to the third is 1 and from the third to the fourth is \displaystyle \frac{3}{4}, and the first term is 0, so the series is not geometric. It does telescope. Try \displaystyle c_n = \frac{n}{2^n}:

c_n - c_{n+1} = \frac{2n - (n+1)}{2^{n+1}} = \frac{n-1}{2^{n+1}}

So \displaystyle S_N = c_1 - c_{N+1} = \frac{1}{2} - \frac{N+1}{2^{N+1}}. An exponential outgrows any power of N, so \displaystyle \frac{N+1}{2^{N+1}} \to 0 and the sum is \displaystyle \frac{1}{2}. A difference of shifted copies is the mark of telescoping; a constant ratio is the mark of a geometric series. The first series has both marks, the second only one.

Example 7

A geometric series as the yardstick

Show that \displaystyle \displaystyle\sum_{n=1}^{\infty} \frac{2^n}{5^n + 3} converges, and bound its sum.

Solution

The series is not geometric: the +3 spoils the constant ratio. But each term is positive and smaller than a geometric term, because making the bottom smaller makes the fraction larger:

0 < \frac{2^n}{5^n + 3} < \frac{2^n}{5^n} = \left(\frac{2}{5}\right)^n

The partial sums of the given series increase, since every term is positive, and each one is less than the matching partial sum of \displaystyle \sum \left(\frac{2}{5}\right)^n, which in turn is less than that series’ sum \displaystyle \frac{2/5}{3/5} = \frac{2}{3}. An increasing sequence that is bounded above converges, so the series converges, with sum less than \displaystyle \frac{2}{3}. Adding terms gives about 0.4979.

This is the pattern of the comparison tests to come: a series is judged by a geometric series (or a p-series) it resembles. Here the ratio of consecutive terms tends to \displaystyle \frac{2}{5}, and the The ratio and root tests turns exactly that observation into a test.

Boundaries

\lvert r \rvert \ge 1 means divergence, including r = -1. For r = -1 the partial sums hop between a and 0; for r = 1 they are Na. Neither has a limit when a \ne 0.
The ratio must be the same for every term from the start. If b_{n+1}/b_n depends on n (Example 2(c), Example 6), the formula \displaystyle \frac{a}{1-r} does not apply. A series that becomes geometric only from some index on is still decided by its ratio, but its sum is the finitely many early terms plus the geometric tail.
The formula needs the actual first term. \displaystyle \frac{a}{1-r} uses the term at the starting index. The constant C in C d^{\,n} is the first term only when the series starts at n = 0.
Splitting a sum needs both pieces to converge. \sum (b_n + c_n) = \sum b_n + \sum c_n holds when both series on the right converge. Two divergent geometric series with the same ratio can add to a convergent one: 2^n + (-2^n) = 0.
Only exponents linear in n give a constant ratio. Terms such as 2^{-n^2} or \displaystyle n\left(\frac{1}{2}\right)^n are not geometric, and the test says nothing about them. Other tests, often by comparison with a geometric series, decide them.

Where this goes next

Geometric series are one of two families of benchmark series; the other is the p-series of The integral test and p series. The Convergence tests judge a series by comparing it with one of them: the The direct and limit comparison tests asks whether the terms behave like a known series’ terms in the long run, and the The ratio and root tests asks whether consecutive terms shrink by a ratio that settles below 1, as if the series were geometric from far out. Read with x in place of r, the sum \displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} for \lvert x \rvert < 1 is a function written as a series, the starting point of Representing functions as power series.

Exercises

Reading off the first term and the ratio

Decide whether each series converges. If it does, find its sum.

\sum_{n=1}^{\infty} \dfrac{3^{n+2}}{4^n}
\sum_{n=0}^{\infty} \dfrac{2^{3n}}{9^{n+1}}
\sum_{n=1}^{\infty} \dfrac{(-3)^{n-1}}{5^n}
\sum_{n=1}^{\infty} \dfrac{6^n}{5^{n+1}}
\sum_{n=0}^{\infty} e^{-n/2}
\sum_{n=1}^{\infty} \dfrac{(-1)^n\, 2^{2n-1}}{7^n}
\sum_{n=0}^{\infty} \dfrac{3^n}{\pi^{n-1}}
\sum_{n=1}^{\infty} 10(0.3)^{2n}

Other starting indices

\sum_{n=5}^{\infty} 3\left(\dfrac{3}{4}\right)^n
\sum_{n=3}^{\infty} \dfrac{5 \cdot 2^n}{3^{n+1}}
\sum_{n=-2}^{\infty} \left(\dfrac{2}{5}\right)^n

Sums of two geometric series

\sum_{n=0}^{\infty} \dfrac{3^n + 4^n}{7^n}
\sum_{n=1}^{\infty} \dfrac{5^n - 2^n}{6^n}
\sum_{n=1}^{\infty} \dfrac{2^n + 7^n}{6^n}

Telescoping and geometric

Find \sum_{n=1}^{\infty} \left(5^{-n} - 5^{-n-1}\right) by telescoping and again as a geometric series.
Show that \sum_{n=1}^{\infty} \dfrac{2n+1}{3^{n+1}} is not geometric, then find its sum by writing the term as c_n - c_{n+1} with c_n = \dfrac{n+1}{3^n}.
Find \sum_{n=1}^{\infty} \left(2^{-n} - 3^{-n}\right). As written, the term is not a difference of shifted copies of one sequence. Find c_n with 2^{-n} - 3^{-n} = c_n - c_{n+1} and c_n \to 0 anyway, and check that c_1 is the sum.

A setting, and the yardstick

Ganzorig strikes a lower string that holds 80 millijoules and loses 1.5\% of its energy each cycle, so it gives off 1.2(0.985)^{n-1} millijoules in cycle n. Find the total energy given off, and the energy given off from cycle 41 on, to two decimal places.
Show that \sum_{n=1}^{\infty} \dfrac{1}{4^n + n^2} converges with sum less than \displaystyle \frac{1}{3}.

Reasoning

Show that every geometric series \sum C d^{\,n} with d \ne 1 telescopes, by finding c_n with C d^{\,n} = c_n - c_{n+1}. Why does this fail for d = 1?
Suppose every b_n \ne 0 and \sum_{n=1}^{\infty} b_n and \sum_{n=1}^{\infty} b_n^2 are both geometric. Show that if the first converges, so does the second, and find the ratio of the second.
Khulan claims that adding two divergent geometric series term by term always gives a divergent series. Give a counterexample, and say which hypothesis of the result on two geometric series it breaks.
Show that \sum_{n=1}^{\infty} \dfrac{3n}{4^n} is not geometric, then use n \le 2^n to show that it converges with sum at most 3.

Answers

Converges to 27. The term is \displaystyle 9\left(\frac{3}{4}\right)^n, first term \displaystyle \frac{27}{4}, so the sum is \displaystyle \frac{27/4}{1/4}.
Converges to 1. 2^{3n} = 8^n and 9^{n+1} = 9 \cdot 9^n, so the term is \displaystyle \frac{1}{9}\left(\frac{8}{9}\right)^n and the sum is \displaystyle \frac{1/9}{1/9}.
Converges to \displaystyle \frac{1}{8}. The term is \displaystyle \frac{1}{5}\left(-\frac{3}{5}\right)^{n-1}, first term \displaystyle \frac{1}{5}, sum \displaystyle \frac{1/5}{8/5}.
Diverges. The term is \displaystyle \frac{1}{5}\left(\frac{6}{5}\right)^n, ratio \displaystyle \frac{6}{5} > 1.
Converges to \displaystyle \frac{1}{1 - e^{-1/2}} = \frac{\sqrt{e}}{\sqrt{e} - 1} \approx 2.5415. The ratio is e^{-1/2} \approx 0.607 and the first term is 1.
Converges to \displaystyle -\frac{2}{11}. The term is \displaystyle \frac{1}{2}\left(-\frac{4}{7}\right)^n, first term \displaystyle -\frac{2}{7}, sum \displaystyle \frac{-2/7}{11/7}.
Converges to \displaystyle \frac{\pi^2}{\pi - 3} \approx 69.70. The term is \displaystyle \pi\left(\frac{3}{\pi}\right)^n, ratio \displaystyle \frac{3}{\pi} \approx 0.955 < 1, first term \pi, sum \displaystyle \frac{\pi}{1 - 3/\pi}.
Converges to \displaystyle \frac{90}{91}. (0.3)^{2n} = (0.09)^n, first term 0.9, sum \displaystyle \frac{0.9}{0.91}.
\displaystyle \frac{729}{256}. First term \displaystyle 3\left(\frac{3}{4}\right)^5 = \frac{729}{1024}, ratio \displaystyle \frac{3}{4}, sum \displaystyle \frac{729}{1024} \cdot 4.
\displaystyle \frac{40}{27}. First term \displaystyle \frac{5 \cdot 8}{81} = \frac{40}{81}, ratio \displaystyle \frac{2}{3}, sum \displaystyle \frac{40}{81} \cdot 3.
\displaystyle \frac{125}{12}. First term \displaystyle \left(\frac{2}{5}\right)^{-2} = \frac{25}{4}, ratio \displaystyle \frac{2}{5}, sum \displaystyle \frac{25/4}{3/5}.
\displaystyle \frac{49}{12}. The pieces \displaystyle \left(\frac{3}{7}\right)^n and \displaystyle \left(\frac{4}{7}\right)^n sum to \displaystyle \frac{7}{4} and \displaystyle \frac{7}{3}.
\displaystyle \frac{9}{2}. From n = 1, \displaystyle \sum \left(\frac{5}{6}\right)^n = 5 and \displaystyle \sum \left(\frac{1}{3}\right)^n = \frac{1}{2}.
Diverges. \displaystyle \left(\frac{1}{3}\right)^n gives a convergent piece and \displaystyle \left(\frac{7}{6}\right)^n a divergent one, so the sum diverges.
\displaystyle \frac{1}{5}. Telescoping: \displaystyle S_N = 5^{-1} - 5^{-N-1} \to \frac{1}{5}. Geometric: the term is \displaystyle \frac{4}{5}\left(\frac{1}{5}\right)^n, first term \displaystyle \frac{4}{25}, sum \displaystyle \frac{4/25}{4/5}.
\displaystyle \frac{2}{3}. The terms \displaystyle \frac{1}{3}, \frac{5}{27}, \frac{7}{81} have ratios \displaystyle \frac{5}{9} and \displaystyle \frac{7}{15}, so it is not geometric. \displaystyle c_n - c_{n+1} = \frac{3(n+1) - (n+2)}{3^{n+1}} = \frac{2n+1}{3^{n+1}}, so \displaystyle S_N = c_1 - c_{N+1} \to c_1 = \frac{2}{3}.
\displaystyle \frac{1}{2}. Both pieces converge: \sum 2^{-n} = 1 and \displaystyle \sum 3^{-n} = \frac{1}{2}. Applying the telescoping result to each piece gives \displaystyle c_n = 2^{1-n} - \frac{1}{2} \cdot 3^{1-n}, since 2^{1-n} - 2^{-n} = 2^{-n} and \displaystyle \frac{1}{2}\left(3^{1-n} - 3^{-n}\right) = 3^{-n}. Then c_n \to 0 and \displaystyle c_1 = 1 - \frac{1}{2} = \frac{1}{2}, the same sum: a sum of two geometric series telescopes too, though not in the form it was written.
80 millijoules in all, and about 43.71 millijoules from cycle 41 on. The total is \displaystyle \frac{1.2}{0.015}; the tail has first term 1.2(0.985)^{40} and sum 80(0.985)^{40}.
\displaystyle 0 < \frac{1}{4^n + n^2} < \frac{1}{4^n}, so the partial sums increase and stay below \displaystyle \sum \left(\frac{1}{4}\right)^n = \frac{1/4}{3/4} = \frac{1}{3}. The sum is about 0.2687.
\displaystyle c_n = \frac{C d^{\,n}}{1 - d} works: \displaystyle c_n - c_{n+1} = \frac{C d^{\,n}(1 - d)}{1 - d} = C d^{\,n}. For d = 1 the term is the constant C, and c_n - c_{n+1} = C for every n forces c_n = c_1 - (n-1)C, which has no limit when C \ne 0, so the telescoping gives no sum.
If b_{n+1} = r b_n, then b_{n+1}^2 = r^2 b_n^2, so the second series has ratio r^2. The first converges, so \lvert r \rvert < 1, hence r^2 < 1 and the second converges, to \displaystyle \frac{b_1^2}{1 - r^2}.
\sum 2^n and \sum \left(-2^n\right) both diverge, but adding them gives terms 0 and the sum 0. The result needs different ratios, p \ne q; here both ratios are 2.
The ratio \displaystyle \frac{3(n+1)/4^{n+1}}{3n/4^n} = \frac{n+1}{4n} depends on n. Since n \le 2^n, \displaystyle 0 < \frac{3n}{4^n} \le 3\left(\frac{1}{2}\right)^n, so the increasing partial sums stay at most \displaystyle 3 \cdot \frac{1/2}{1/2} = 3, and the series converges. (Its sum is \displaystyle \frac{4}{3}.)