Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find whether the series diverges and its sum: $$\sum_{n = 1}^\infty (-1)^{n+1} \frac{3}{5^n}.$$

I found that the series converges using the Alternating Series test because the absolute value of each $n$ decreases while the value of $n$ increases. Then I took the limit as $x$ approaches infinity of $3/5^x$ which is $0$.

But I am not sure how to go about finding the sum at this point. This is a practice exam, so the solution is more important then the actual answer.

share|cite|improve this question
Hint: it's a Geometric series. – David Mitra Dec 8 '12 at 21:02

3 Answers 3

up vote 7 down vote accepted

Notice that$$(-1)^{n+1}\frac{3}{5^n}=-3\frac{(-1)^{n}}{5^{n}}=-3\left(\frac{-1}{5}\right)^{n}$$ Since $\sum_{k=1}^{\infty}ar^{k}=\frac{ar}{1-r}$ (iff $|r|<1$), $$\sum_{n=1}^{\infty}-3\left(\frac{-1}{5}\right)^{n}=\frac{-3\cdot\frac{-1}{5}}{1-\frac{-1}{5}}=\frac{\frac{3}{5}}{\frac{6}{5}}=\frac{1}{2}$$ and the sum converges because $\left|\frac{-1}{5}\right|=\frac{1}{5}<1$

share|cite|improve this answer

Hint: Use the following

$$ \sum_{k=0}^{\infty}(-1)^kx^k=\frac{1}{1+x}. $$

share|cite|improve this answer

Your series is an example of a geometric series. The first term is $a=3/5$, while each subsequent term is found by multiplying the previous term by the common ratio $r=-1/5$.

There is a well known formula for the sum to infinity of a geometric series with $|r| < 1$, namely:

$$S_{\infty} = \frac{a}{1-r} \, . $$

In your case, $a=3/5$ and $r=-1/5$, and so it follows that:

$$S_{\infty} = \frac{3/5}{1+1/5} = \frac{1}{2} \, . $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.