Question

Determine whether the series converges or diverges. Verify that the test used is applicable:

I have
a)
\[\sum_{n=2}^{\infty} \frac{121+\sin(121n)}{n^{21} + 21n}\]
b)
\[\sum_{n=4}^{\infty}\frac{2012}{\ln(n^{n})}\]
a)
I split the series and used the comparison test \[\frac{121}{n^{21}+21n}\le \frac{121}{n^{21}}\]
\[\frac{\sin(121n)}{n^{21}+21n} \le \frac{1}{n^{21}}\]
by p test they both converge

b)
I used the integral test, determined
-the series is decreasing
\[\frac{2012}{\ln(n^{n})} \ge \frac{2012}{\ln(n+1)^{n+1}}\]
-the function is decreasing
\[f(x) = -\frac{2012}{\ln(n^{

Answer 1

-the function is decreasing
-2012/ln(n^(n))^(2)
- and the function is always positive

thus I applied the integral test

\[\int\limits_{\infty}^{4}\frac{2012}{\ln(n^{n})} = 2012\ln|n^{n}|\left(\begin{matrix}\infty \\ 4\end{matrix}\right)\]

Answer 2

please ignore the brackets I don't know how to use proper integral notation on this site, anyways, I got that the integral was equal to infinity and thus divergent

therefore the series is divergent by integral test.


My main question is did I do this correctly and how do I verify that the test used is applicable?

Answer 3

wait I made a mistake solving the integral of 121/ln(x^(x))

Answer 4

u = ln(x) + 1
du/(ln(x)+1) = dx

Answer 5

this integral is not easily solved ugh

Answer 6

nvm I will just use partial fractions :)