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Prove the series $$\sum _{n=1}^{\infty }\left(-1\right)^{n-1}\frac{2n+1}{n\left(n+1\right)}\:$$ converges, and prove it's sum is $1$.

So I proved it's converging conditionally, and I'm having troubles with the sum.

From Wolfram Alpha I can see what the partial sum formula is, but I can't get to it.

Obviously this is a telescoping series, and I did get that $\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}$, but not sure what to do with that.

Any help or hints appreciated.

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  • $\begingroup$ use induction to pove the formula of the finite series $\endgroup$ – Dr. Sonnhard Graubner Feb 24 '17 at 16:35
  • $\begingroup$ Try expanding the fraction into its two constituent parts and then employ the partial fraction decomposition you stated above. I.e. $(-1)^{n-1}\left( \frac{2n}{n(n+1)} + \frac{1}{n(n+1)}\right)$ $\endgroup$ – FofX Feb 24 '17 at 16:39
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$$\sum_{r=1}^n(-1)^{r-1}\left(\dfrac1r+\dfrac1{r+1}\right)$$

$$=\dfrac11+\dfrac12-\dfrac12-\dfrac13+\dfrac13+\dfrac14-\dfrac14+\cdots+(-1)^{n-2}\left(\dfrac1{n-1}+\dfrac1n\right)+(-1)^{n-1}\left(\dfrac1n+\dfrac1{n+1}\right)$$

$$=1+(-1)^{n-1}\dfrac1{n+1}$$

the two surviving terms are the first part of the first summand and the last part of the last summand

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  • $\begingroup$ Could you please explain how did you get $\left(\frac{1}{r}+\frac{1}{r+1}\right)$? $\endgroup$ – Itay4 Feb 24 '17 at 16:40
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    $\begingroup$ $$\dfrac{2r+1}{r(r+1)}=\dfrac{r+(r+1)}{r(r+1)}=?$$ $\endgroup$ – lab bhattacharjee Feb 24 '17 at 16:41
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HINT: $$\frac{2n + 1}{n(n+1)} = \frac{2}{n+1} + \frac{1}{n} - \frac{1}{n+1} = \frac{1}{n} + \frac{1}{n+1} $$

Now use the alternating sign to cancel almost all of the terms.

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