# infinite series--integral?

• March 21st 2010, 11:31 PM
234578
infinite series--integral?
Hi, I am supposed to determine whether "the sum of the sequence a(n)from n=1 to n=infinity" converges (i.e. whether or not the sequence is summable) where a(n)=1/[n^(1+(1/n)]. I figure I can use the fact that this sequence is summable if lim(A->infinity)[integral from 1 to A of f(x)] exists where f(x)=1/[x^(1+(1/x))]. My problem is I forget (or maybe never knew) how to find this integral which I must do to find the limit of the integral as A approaches infinity. Am I even approaching the question in the correct way?

Thankyou for any help
• March 22nd 2010, 12:03 AM
chisigma
Once You have the explicit expressione for the $a_{n}$, finding the integral for the covergence test is very easy: You have only to swap the $n$ and the $x$...

$a_{n} = \frac{1}{n^{1+\frac{1}{n}}} \rightarrow f(x)= \frac{1}{x^{1+\frac{1}{x}}}$ (1)

Kind regards

$\chi$ $\sigma$
• March 22nd 2010, 07:11 AM
234578
Quote:

Originally Posted by chisigma
Once You have the explicit expressione for the $a_{n}$, finding the integral for the covergence test is very easy: You have only to swap the $n$ and the $x$...

$a_{n} = \frac{1}{n^{1+\frac{1}{n}}} \rightarrow f(x)= \frac{1}{x^{1+\frac{1}{x}}}$ (1)
$\chi$ $\sigma$
Thankyou for assuring me that I am going in the right direction with the integral test. I know I have to swap the n and the x and find the integral of $f(x)= \frac{1}{x^{1+\frac{1}{x}}}$ (1) but my question is how do I do that? I can't use integration by parts because logx is missing from the equation (the derivative of 1+ 1/x would be logx right?) so I don't know how to integrate this.