# How to decide convergence or otherwise of these series?

How to decide whether the following three series converge or diverge?

1. $$\sum_{n=2}^{\infty} \frac{(-1)^n}{\sqrt{n} + (-1)^n}$$ By the limit comparison with the divergent series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ we can say that this series is not absolutely convergent. However since $$\frac{1}{\sqrt{n}+(-1)^n} - \frac{1}{\sqrt{n+1}+(-1)^{n+1}} = \frac{ \sqrt{n+1} - \sqrt{n} + 2 (-1)^{n+1}}{(\sqrt{n}+(-1)^n)(\sqrt{n+1}+(-1)^{n+1})},$$ which is positive if $n$ is odd but negative if $n$ is even. So we can't apply the Leibnitz's rule.

2. $$\sum_{n=1}^{\infty} \frac{1}{\log ( e^n + e^{-n} ) }$$ By the Leibnitz's rule we can conclude that the corresponding alternating series converges (conditionally).

3. $$\sum_{n=1}^{\infty} (-1)^n \int_{n}^{n+1} \frac{e^{-x}}{x} \ dx$$ The integral here satisfies the inequalities $$\frac{e^{-n}-e^{-n-1}}{n+1} \leq \int_{n}^{n+1} \frac{e^{-x}}{x} \ dx \leq \frac{e^{-n}-e^{-n-1}}{n},$$ and both the upper and lower bounds approach $0$ as $n$ increases without bound. So we can't apply the $n$-th term test for divergence.

For the first problem, add together terms $2$ and $3$, terms $4$ and $5$, and so on. If $n$ is even, the sum of the $n$-th and $(n+1)$-th terms ia $$\frac{\sqrt{n+1}-\sqrt{n}-2}{(\sqrt{n}+1)(\sqrt{n+1}-1)}.$$ This can be rewritten as $$\frac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n}+1)(\sqrt{n+1}-1)}-\frac{2}{(\sqrt{n}+1)(\sqrt{n+1}-1)}.\tag{1}$$

The sum of the first parts of (1) converges. To show this, rationalize the numerator.

The sum of the second parts of (1) diverges, essentially by comparison with the harmonic series. So the original series diverges.

Remark: If you need to be more formal, consider the sum from $n=2$ to $n=2m+1$. The sum of the first parts of (1) is bounded, and the sum of the second parts is not. So partial sums do not converge.

The first sum is divergent:

Let $a_n = \frac{(-1)^n}{\sqrt{n} + (-1)^n}$. Consider $a_n + a_{n+1}$ for even values of $n$. We have $$a_n + a_{n+1} = \frac{1}{\sqrt{n} + 1} - \frac{1}{\sqrt{n+1} - 1} = \frac{\sqrt{n+1} - \sqrt{n} - 2}{(\sqrt{n} + 1)(\sqrt{n+1} - 1)}$$ The nominator is smaller that $-1$ for all $n \geq 2$. The denominator is larger than $n/2$ for all $n \geq 2$. Therefore $a_n + a_{n+1} \leq -\frac{2}{n}$ for all even $n \geq 2$. Therefore, $$\sum_{n=2}^\infty a_n = - \infty$$

The second one is divergent: $$\sum_{n=1}^\infty \frac{1}{\log(e^n + e^{-n})} \geq \sum_{n=1}^\infty \frac{1}{\log(e \cdot e^n)} = \sum_{n=1}^\infty \frac{1}{1 + n} \geq \sum_{n=1}^\infty \frac{1}{2n} = \infty.$$

The third one is absolutely convergent: $$\sum_{n=1}^\infty \left| (-1)^n \int_{n}^{n+1} \frac{e^{-x}}{x} \right| \leq \sum_{n=1}^\infty e^{-n} < \infty$$

• As I've demonstrated in my question itself, the first series can't be dealt with using the alternating series test as the function in question doesn't happen to be a decreasing one. – Saaqib Mahmood Aug 18 '13 at 12:20
• @upvoter Why the upvote? – Did Aug 18 '13 at 13:06
• @Saaqib sorry, you are right of course. I fixed the answer. Hope it is correct now. – Igor Shinkar Aug 18 '13 at 14:04