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If $|a_{n}-a_{m}|\leq \frac{1}{\min{(m, n)}}$ then $\sum_{n=1}^{\infty}(a_{n}-a_{n+2})$ converges

Using the fact that $|a_{n}-a_{m}|\leq \frac{1}{\min{(m, n)}}$ I showed that $a_{n}$ is Cauchy hence $a_{n}\to L$.

If we let $b_{n}=a_{n}-a_{n+2}$ then by looking at the partial sums: $$S_{2n}=\sum_{k=1}^{2n}b_{k}=a_{2}-a_{2n}\to a_{2}-L$$ and $$S_{2n-1}=\sum_{k=1}^{2n-1}b_{k}=a_{1}-a_{2n-1}\to a_{1}-L$$

But I seem to missing something because if $a_{1}\neq a_{2}$ the above partial sums don't converge to the same limit and the series diverges. Where is the error?

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+1 for showing your work – Arturo Magidin Feb 7 '11 at 18:08
$S_n = a_1 + a_2 - a_{n+1} - a_{n+2}$. – user17762 Feb 7 '11 at 18:09
up vote 2 down vote accepted

Your formulas for $S_{2n}$ and for $S_{2n-1}$ are incorrect.

For example, take $n=2$. You claim that $S_4 = a_2 - a_4$, but in fact, \begin{align*} S_4 &=\mathop{\sum}\limits_{k=1}^4b_k = (a_1-a_3) + (a_2-a_4) + (a_3-a_5) + (a_4-a_6)\\ &= a_1+a_2 - (a_5+a_6). \end{align*} And you claim $S_3=a_1 - a_3$, when in fact \begin{align*} S_3 &=\mathop{\sum}\limits_{k=1}^3b_k = (a_1-a_3) + (a_2-a_4) + (a_3-a_5)\\ &= a_1+a_2 - (a_4+a_5). \end{align*} Get the right formulas for $S_{2n}$ and $S_{2n-1}$, and things should work out fine.

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that mistake cost me 10 points, why didn't I think of rechecking my calculation! – user6751 Feb 7 '11 at 18:23

There's an error in your calculation of the partial sums, but otherwise you're on the right track. Notice that $$\sum_{n=1}^{N} a_n -a_{n+2} = \sum_{n=1}^{N} a_n - \sum_{m=3}^{N+2} a_m.$$

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