What is the solution to evaluating this fraction series?

[anemone]

Here is this week's POTW:

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Evaluate $\dfrac{7}{1^2\cdot 6^2}+\dfrac{17}{6^2\cdot 11^2}+\dfrac{27}{11^2\cdot 16^2}+\cdots$-----

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[anemone]

Congratulations to the following members for their correct solution::)

1. greg1313
2. Opalg
3. kaliprasad

Solution from Opalg:

Spoiler

We need to evaluate \(\displaystyle \sum_{k=1}^\infty \frac{10k-3}{(5k-4)^2(5k+1)^2}.\)

In fact, \(\displaystyle \sum_{k=1}^n \frac{10k-3}{(5k-4)^2(5k+1)^2} = \frac15\Bigl(1 - \frac1{(5n+1)^2}\Bigr).\)

To prove that by induction, check first that when $n=1$ that formula gives $\frac15\Bigl(1 - \frac1{36}\Bigr) = \frac{36-1}{5\times 36} = \frac7{1^2\cdot6^2}$, which is the first term of the series.

For the inductive step, we have to add the $(n+1)$th term of the series, which is \(\displaystyle \frac{10n+7}{(5n+1)^2(5n+6)^2}.\) So if the sum of the first $n$ terms is \(\displaystyle \frac15\Bigl(1 - \frac1{(5n+1)^2}\Bigr)\) then the sum of the first $n+1$ terms will be $$ \frac15\Bigl(1 - \frac1{(5n+1)^2}\Bigr) + \frac{10n+7}{(5n+1)^2(5n+6)^2} = \frac15\Bigl(1 - \frac1{(5n+1)^2} + \frac{50n+35}{(5n+1)^2(5n+6)^2}\Bigr).$$ But $$ \begin{aligned}- \frac1{(5n+1)^2} + \frac{50n+35}{(5n+1)^2(5n+6)^2} &= \frac{-(5n+6)^2 + 50n+35}{(5n+1)^2(5n+6)^2} \\ &= \frac{-25n^2 - 60n - 36 + 50n + 35}{(5n+1)^2(5n+6)^2} \\ &= \frac{-25n^2 - 10n - 1}{(5n+1)^2(5n+6)^2} \\ &= \frac{-(5n+1)^2}{(5n+1)^2(5n+6)^2} = -\frac1{(5n+6)^2}.\end{aligned}$$ Hence the sum of the first $n+1$ terms of the series is \(\displaystyle \frac15\Bigl(1-\frac1{(5n+6)^2}\Bigr),\) which is what the claimed formula gives.

That completes the inductive proof. Finally, as $n\to\infty$ the formula gives the sum of the whole series as \(\displaystyle \boxed{\frac15}\).