Determine for which $\alpha > 0$ the series $\sum_\limits{n=1}^{\infty}\frac{ne^n-\log(1+n)}{n}(\frac{1}{n^{\alpha}})$ converges

Question

Determine for which $\alpha > 0$ the following series converges $$\sum_\limits{n=1}^{\infty}\frac{ne^n-\log(1+n)}{n}\frac{1}{n^{\alpha}}$$

My attempt:

$$\sum_\limits{n=1}^{\infty}\dfrac{ne^n-\log(1+n)}{n}\dfrac{1}{n^{\alpha}}=\sum_\limits{n=1}^{\infty}\dfrac{ne^n-\log(1+n)}{n^{a+1}}$$ Then I separated the series: (Can I do this?) $$\sum_\limits{n=1}^{\infty}\dfrac{ne^n}{n^{a+1}}-\sum_\limits{n=1}^{\infty}\dfrac{\log(1+n)}{n^{a+1}}$$

So:

I tried ratio test with the first one and it diverges $\forall \alpha>0$.

The second one tends to $\sum_\limits{n=1}^{\infty}\dfrac{\log(n)}{n^{a+1}}$ which converges $\forall \alpha>0$.

So the entire series diverges.

Have I done something wrong?

Answer 1

You are right, the series is divergent. However your argument works only when $\alpha>0$ because in that case the second series is convergent. Otherwise we have an indeterminate form $+\infty-\infty$ and the separation is not allowed.

More simply, for any $\alpha\in \mathbb{R}$, $$a_n:=\dfrac{ne^n-\log(1+n)}{n}\cdot\dfrac{1}{n^{\alpha}}\sim e^n\cdot n^{-\alpha}\to +\infty$$ which implies that $\sum_n a_n=+\infty$.

Answer 2

$$ne^n-\ln (1+n)\sim ne^n \;\;(n\to+\infty) $$

the general term never goes to zero, the series is always divergent.