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I am required to find the values of $x$ in the following infinite series, which cause the series to converge. $$\sum_{n=1}^\infty \frac{x^n}{\ln(n+1)}$$

I tried to use the ratio test, and found that the series converges when $x$ is in $(-1,1)$. However, this answer is not correct.

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You also need to consider $x=1$ (which does not converge) and $x=-1$, which is convergent, using Abel's test. – user77893 May 15 '13 at 6:53
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You are largely right when you say that the series converges for all $x$ in the interval $(-1,1)$. However, the series also converges for $x=-1$, because then it is an alternating series.

The series diverges at $x=1$, and also whenver $|x|\gt 1$.

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From ratio test, as you have concluded, we need $$\limsup_{n \to \infty} \left \vert\dfrac{x^{n+1}/\ln(n+2)}{x^n/\ln(n+1)}\right \vert < 1$$ Hence, we need $$\vert x \vert \limsup_{n \to \infty} \left \vert\dfrac{\ln(n+1)}{\ln(n+2)}\right \vert < 1 \implies \vert x \vert < 1$$ We need to check the boundaries. For $x=-1$, this is an alternating series and hence converges. At the other boundary, i.e., $x=1$, it clearly diverges. Hence, converges for $x \in [-1,1)$. If you are considering $x \in \mathbb{C}$, we then have that the series converges for all $x$ such that $\vert x \vert \leq 1$ and $x \neq 1$. The convergence on the boundary (not including $1$), can be proved using the Dirichlet test/ generalized alternating test.

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