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<a_{n}=2^{\frac{1}{n}}-2^{\frac{1}{n+1}} < 2^{\frac{1}{n}} = b_{n} \end{align*} 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!

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

2 Answers 2


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.


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