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enter image description here It's a simple question . If the value of x is 1 then the series becomes summation of 1/n which is divergent. But in the answer it is given that the series converges. I think the answer is wrong.

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    $\begingroup$ The given series is absolutely convergent for any $x\in(-1,1)$ and conditionally convergent for any $x\in[-1,1)$. $\endgroup$ – Jack D'Aurizio Feb 1 '17 at 2:46
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Using the libnitz test you can find out that the radius of convergence is (-1,1) and hence it won't convergence at 1.

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The answer is wrong since it converges for $x\in[-1,1)$ but it does not converge at $x=1$.

Applying the ratio test we see that it converges when

$$ \lim_{x\to\infty}\left\lvert \frac{x^{n+1}}{n+1}\cdot\frac{n}{x^n}\right\rvert<1$$

But

$$\lim_{x\to\infty}\left\lvert \frac{x^{n+1}}{n+1}\cdot\frac{n}{x^n}\right\rvert=\vert x\vert<1$$

so the series converges on $(-1,1)$. For $x=-1$ it is the alternating harmonic series which converges. At $x=1$ it is the harmonic series which diverges.

So the series converges for $x\in[-1,1)$.

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