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We have $\sum^{\infty}_{n=0} \frac{x^{2n+1}}{2n+1}$

The radius of convergence is 1 (ratio test)

How about the convergence for -1 and 1 ?

Let's say x=1 , then we have $\sum^{\infty}_{n=0} \frac{1}{2n+1}$

Does it converge? I guess not, because we are summing up a sequence which will never be a zero sequence. So for x = 1 the series is not converge.

And for x = -1 we have $\sum^{\infty}_{n=0} \frac{-1}{2n+1}$ which is also not a zero sequence. So the series diverges for x=1 and x=-1

Is the way I think right?

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    What about Taylor series for this sum ? – 2017-01-25
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    I am confused enough, please let put taylor on the side :D Would be nice if my calculus here is right. – 2017-01-25
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    "We are summing up a sequence which will never be a zero sequence" is correct,but the conclusion you draw from it is just plain wrong. There are plenty of sequences that are never zero, but have finite sum. For instance, you get one if you insert $z = 1/2$ into your series. Instead, you have to compare your series with series whose sum is _known_ to diverge or converge. – 2017-01-25

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For $x=1$, we have: $\sum^{\infty}_{n=0} \frac{x^{2n+1}}{2n+1}$. Now for $n\geq2$ we have: $2n+1 < 3n$, so: $\frac{1}{2n+1}>\frac{1}{3n}$. Now as we know that $\sum^{\infty}_{n=0} \frac{1}{3n}$ diverges, what does the comparison test then tell us about $\sum^{\infty}_{n=0} \frac{x^{2n+1}}{2n+1}$?

For $x=-1$ the series also diverges (to $-\infty$), because of the presence of the minus sign combined with the comparison test.

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    Wolfram-Alpha says, that the series diverges for x=-1, why? – 2017-01-25
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    Oops, excuse me. I thought there was a $(-1)^n$ in the numerator, which clearly is not there. For $x=-1$ the series indeed diverges to $-\infty$! – 2017-01-25
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    Do I have to make limes test after I do x= -1? Because $(-1)^{2n+1}$ is always (-1), so we have $ \frac{-1}{2n+1} $ Why does that diverges to $ - \infty $ ? – 2017-01-25
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    You can use the comparison test again. Or simply prove that if $\sum a_n$ (with $a_n>0$ for all $n$) diverges to $+\infty$, then $\sum - a_n$ diverges to $-\infty$. – 2017-01-25
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    I see, thank you very much, Helios! – 2017-01-25