0
$\begingroup$

I want to check, whether $\sum\limits_{n=0}^{\infty}\frac{1}{n+\exp(n)}$ converges or diverge.

I tried to use the comparison test: $$|\frac{1}{n+\exp(n)}|\leq \frac{1}{\exp(n)} = \frac{1}{\frac{n^n}{n!}} = \frac{n!}{n^n} $$

$\frac{n!}{n^n}$ converge $\Rightarrow$ $\sum\limits_{n=0}^{\infty}\frac{1}{n+\exp(n)}$ converge.

Can somebody please tell me, if this is correct?

$\endgroup$
  • $\begingroup$ Do you think that $\exp(n) = \frac{n^n}{n!}$? $\endgroup$ – Najib Idrissi Dec 6 '13 at 19:09
  • $\begingroup$ $\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}$ is the definition of $\exp(x)$. So $\exp(n) = \frac{n^n}{n!}$. Can somebody please tell me, if this is correct? $\endgroup$ – fear.xD Dec 6 '13 at 19:20
  • 2
    $\begingroup$ No, this is not correct. You cannot reuse the index in the sum. $$\exp(n) = \sum_{k=0}^\infty \frac{n^k}{k!}$$ $\endgroup$ – Najib Idrissi Dec 6 '13 at 19:28
8
$\begingroup$

There is a much easier way. Note that $$\dfrac1{n+e^n} < \dfrac1{e^n}$$ Hence, $$\sum_{n=0}^{\infty}\dfrac1{n+e^n} < \sum_{n=0}^{\infty}\dfrac1{e^n} = \dfrac{e}{e-1}$$

$\endgroup$
1
$\begingroup$

Use Quotient test to see that $$n^{3/2}u_n\to 0<\infty, ~~n\to\infty$$

$\endgroup$
  • $\begingroup$ It is clear that $u_n=O(n^2)$ $\endgroup$ – Gabriel Romon Dec 6 '13 at 19:19
  • $\begingroup$ @GabrielR.: So? :) $\endgroup$ – mrs Dec 6 '13 at 19:27
  • $\begingroup$ I forgot the minus sign $O(n^{-2} )$. I like the bigO notation better, as everybody :) . $\endgroup$ – Gabriel Romon Dec 6 '13 at 19:35
0
$\begingroup$

The transition $1/e^n = \frac{n^n}{n!} $ inaccurately and needs some explanation, what you can do instead is use the fact the $a_n=1/e^n$ is geometric sequence.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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