# Does $\sum \limits_{n=1}^{\infty} \frac{1}{\ln(n^n+n^2)}$ converge?

I think it does but I'm having trouble showing it.

• Using the integral test: the function $f(x)=\frac{1}{\ln(x^x+x^2)}$ is decreasing but I'm having trouble integrating $\int_{1}^{\infty} \frac{dx}{\ln(x^x+x^2)}$, trying the substitution $t=x^x$ didn't seem to work.

• The root test gives $\sqrt[n]{\left |\frac{1}{\ln(n^n+n^2)}\right |} \to 1$ which doesn't imply anything.

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You should be trying a limit comparison. First show that you can ignore the n^2. (The way to use the integral test is generally not to integrate the function itself, but to integrate obvious lower and/or upper bounds on the function which are easier to integrate.) –  Qiaochu Yuan Jan 10 '11 at 19:59

Hint: Compare with $$\sum_{n=1}^{\infty} \frac{K}{n\log n}$$
If my calculations are correct then $K=1$ since: $$1 \leftarrow \frac{n\ln n}{\ln(2n^n)} \leq \frac{n\ln n}{\ln(n^n+n^2)} \leq \frac{n\ln n}{\ln(n^n)}=1$$ And since $\sum_{n=1}^{\infty} \frac{1}{n\log n}$ diverges, the mentioned series diverges. –  daniel.jackson Jan 10 '11 at 20:40
@Daniel: I was think more along the lines of $\frac{1}{\ln (n + n^2)} > \frac{1}{2(n \ln n)}$, but there are too many possibilities. What you have is correct too, in fact you have shown that any $0 < K \lt 1$ can be chosen for sufficiently large $n$. –  Aryabhata Jan 10 '11 at 20:51
@Moron: Yes but if I want to use the limit comparison then I need to show that $\frac{n\ln n}{\ln(n^n+n^2)} \to A \gt 0$, no? –  daniel.jackson Jan 10 '11 at 20:51