# Determine if the following series converges or diverges. I did this...

Determine if this series converges or diverges. Explain why it converges or diverges:

1. $$\sum_{n=1}^{\infty}\left ( 2^{\frac{1}{n}}-2^{\frac{1}{n+1}} \right )$$
2. $$\sum_{n=1}^{\infty}\frac{5^{n}-2^{n}}{7^{n}-6^{n}}$$
3. $$\sum_{n=1}^{\infty}\left ( -1 \right )^{n} \cos({\frac{\pi}{n}})$$

What I've done in 1: Observe that \begin{align*} 0 So, if I prove that $$\sum_{n=1}^{\infty}2^{\frac{1}{n}}$$ converges, then $$\sum_{n=1}^{\infty}\left ( 2^{\frac{1}{n}}-2^{\frac{1}{n+1}} \right )$$ converges. To see that I defined $$b_{n}={2^{\frac{1}{n}}}$$, and then I used the ratio test, this is:

\begin{align*} \lim_{n \rightarrow \infty} \frac{b_{n+1}}{b_{n}}=\lim_{n \rightarrow \infty}\frac{2^{\frac{1}{n+1}}}{2^{\frac{1}{n}}}=\lim_{n \rightarrow \infty}\frac{1}{2^{\frac{1}{n(n+1)}}}=1 \end{align*} So, I can't conclude if $$\sum_{n=1}^{\infty} b_{n}$$ converges or not, even more I can't conclude if $$\sum_{n=1}^{\infty} a_{n}$$ converges or not. Is there any way that I can determine it?

What I've done in 2: With a similar idea than in 1, we have that \begin{align*} 0 < a_{n} = \frac{5^{n}-2^{n}}{7^{n}-6^{n}} < \frac{5^{n}}{7^{n}-6^{n}} = b_{n} \end{align*} With the same idea I used the ratio test: \begin{align*} \lim_{n \rightarrow \infty} \frac{b_{n+1}}{b_{n}}=\lim_{n \rightarrow \infty} \frac{\frac{5^{n+1}}{7^{n+1}-6^{n+1}} }{\frac{5^{n}}{7^{n}-6^{n}} }=\lim_{n \rightarrow \infty} 5\left ( \frac{7^{n}-6^{n}}{7^{n+1}-6^{n+1}} \right )=\frac{5}{7}<1 \end{align*}

Therefore, $$\sum_{n=1}^{\infty} b_{n}$$ converges $$\Rightarrow \sum_{n=1}^{\infty} a_{n}$$ converges. (Is it correct my conclusion?)

For 3, intuitively I think it doesn't converges but I don't know how can I show it since most of the tests to verify if $$\Rightarrow \sum_{n=1}^{\infty} a_{n}$$ converges, establishes that $$a_{n}>0$$. How can I prove it converges or not?

I would really appreciate your help!

• @BrianM.Scott I think that is incorrect. Maybe is because the notation: $\frac{2^{\left ( \frac{1}{n+1} \right )}}{2^{\left ( \frac{1}{n} \right )}}=\frac{\sqrt[n+1]{2}}{\sqrt[n]{2}}$ Sep 25, 2020 at 2:28
• Yes, I misread it or suffered a mental hiccup, I’m not sure which. Sep 25, 2020 at 2:29

For 1, the partial sums are $$\sum_{n=1}^{m-1}\left ( 2^{\frac{1}{n}}-2^{\frac{1}{n+1}} \right ) =2-2^{\frac{1}{m}} \to 1$$ since $$2^{\frac{1}{m}} \to 1$$.
Therefore the sum converges to $$1$$.
If the series $$\sum_{n=1}^{+\infty}{a_{n}}$$ converges, then $$\lim_{n\to+\infty}{a_{n}}=0$$. Notice that $$\lim_{n\to+\infty}{|(-1)^n cos(\frac{\pi}{n})|=1}$$, so the series 3 is not converges.