I was asked to compute this series: ∑n=1∞1(2n−1)2 but by using the fact that ∑n=1∞1n2=π26

By direct calculation, we see that ∑n=1∞1n2=∑k=1∞1(2k−1)2+14∑n=1∞1n2 which means ∑k=1∞1(2k−1)2=34∑n=1∞1n2=34π26=π28

Observe that ∑n=1∞1(2n)2=1(2⋅1)2+1(2⋅2)2+1(2⋅3)2…=14(112+122+132+…)=π24(6) But ∑n=1∞1(2n)2+∑n=1∞1(2n−1)2=∑n=1∞1n2 where rearrangement is allowed since we are dealing with non-negative series.

By direct calculation, we see that ∑n=1∞1n2=∑k=1∞1(2k−1)2+14∑n=1∞1n2 which means ∑k=1∞1(2k−1)2=34∑n=1∞1n2=34π26=π28

Observe that ∑n=1∞1(2n)2=1(2⋅1)2+1(2⋅2)2+1(2⋅3)2…=14(112+122+132+…)=π24(6) But ∑n=1∞1(2n)2+∑n=1∞1(2n−1)2=∑n=1∞1n2 where rearrangement is allowed since we are dealing with non-negative series.

Answered: "I was asked to compute this series: ∑n=1∞1(2n−1)2..."

Nicontio1

Answered question

2021-12-23

I was asked to compute this series:

\sum_{n = 1}^{\infty} \frac{1}{{(2 n - 1)}^{2}}

but by using the fact that

\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6}

Answer & Explanation

scomparve5j

Beginner2021-12-24Added 38 answers

By direct calculation, we see that

\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \sum_{k = 1}^{\infty} \frac{1}{{(2 k - 1)}^{2}} + \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}}

which means

\sum_{k = 1}^{\infty} \frac{1}{{(2 k - 1)}^{2}} = \frac{3}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{3}{4} \frac{π^{2}}{6} = \frac{π^{2}}{8}

kalfswors0m

Beginner2021-12-25Added 24 answers

Observe that

\sum_{n = 1}^{\infty} \frac{1}{{(2 n)}^{2}} = \frac{1}{{(2 \cdot 1)}^{2}} + \frac{1}{{(2 \cdot 2)}^{2}} + \frac{1}{{(2 \cdot 3)}^{2}} \dots = \frac{1}{4} (\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots) = \frac{π^{2}}{4 (6)}

But

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

where rearrangement is allowed since we are dealing with non-negative series.

user_27qwe

Skilled2021-12-30Added 375 answers

Very detailed answer. Thanks.

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