Determine if the following series converges or diverges. I did this...
Determine if this series converges or diverges. Explain why it converges or diverges:
- $\sum_{n=1}^{\infty}\left ( 2^{\frac{1}{n}}-2^{\frac{1}{n+1}} \right )$
- $\sum_{n=1}^{\infty}\frac{5^{n}-2^{n}}{7^{n}-6^{n}}$
- $\sum_{n=1}^{\infty}\left ( -1 \right )^{n} \cos({\frac{\pi}{n}})$
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!
$\endgroup$ 22 Answers
$\begingroup$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 $1$.
$\endgroup$ $\begingroup$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.
$\endgroup$
