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Summation of (-1)^n/(n+(-1)^n+1)

Does the series in the attached image converge?

This is my attempt at a solution:

We have:

-1/2 + 1 -1/4 + 1/3 -1/6 + 1/5 -.....

I had this in an exam paper and answered that the series converges because the series:

1 - 1/2 + 1/3 - 1/4 + ...

converges (according to Liebniz).

But I only got 2/10 for the question because rearranging an infinite series can change the result. So how do I prove this series does or does not converge?

Thanks

Re: Summation of (-1)^n/(n+(-1)^n+1)

Quote:

Originally Posted by

**durrrrrrrr** Does the series in the attached image converge?

This is my attempt at a solution:

We have:

-1/2 + 1 -1/4 + 1/3 -1/6 + 1/5 -.....

I had this in an exam paper and answered that the series converges because the series:

1 - 1/2 + 1/3 - 1/4 + ...

converges (according to Liebniz).

But I only got 2/10 for the question because rearranging an infinite series can change the result. So how do I prove this series does or does not converge?

Let $\displaystyle S_n$ denote the sum of the first n terms of the series. Then $\displaystyle S_{2n}$ is the same as the sum of the first 2n terms of the Leibniz series (because there is no problem with rearranging the terms in a finite sum). So $\displaystyle S_{2n}$ converges to the same limit L as the Leibniz series.

Next, the (2n+1)th term in the series is $\displaystyle -\tfrac1{2(n+1)}$, which converges to 0 as $\displaystyle n\to\infty.$ Therefore $\displaystyle S_{2n+1} = S_{2n} -\tfrac1{2(n+1)} \to L+0 = L.$

Finally, since $\displaystyle S_{2n}$ and $\displaystyle S_{2n+1}$ converge to the same limit L, it follows that the whole series converges to L.

Re: Summation of (-1)^n/(n+(-1)^n+1)

Quote:

Originally Posted by

**Opalg** Finally, since $\displaystyle S_{2n}$ and $\displaystyle S_{2n+1}$ converge to the same limit L, it follows that the whole series converges to L.

Thanks, is this last stage true by induction?