"Proving that:32∑k=1∞4k3+k2=π2−6" was answered by students like you

Alexandria Dodd

Answered question

2022-02-25

Proving that:

\frac{3}{2} \sum_{k = 1}^{\infty} \frac{4}{k^{3} + k^{2}} = π^{2} - 6

zakrwinpfo

Beginner2022-02-26Added 4 answers

Equivalently, you are asking for

6 \sum_{k = 1}^{\infty} \frac{1}{k^{3} + k^{2}}

Using your partial fractions decomposition, we can split this into

6 \sum_{k = 1}^{\infty} (\frac{1}{k^{2}} + \frac{1}{k + 1} - \frac{1}{k})

The second and third terms telescope, giving a sum of -1. The first term sums to

\frac{π^{2}}{6}

which is a well-known fact.
Putting this all together, your series sums to

π^{2} - 6

. You can complete this argument.

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