if the series Σ_(n=1) ^∞ (1/n^2 ) converges to k . find the convergence value of Σ


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Question Number 209837 by lmcp1203 last updated on 23/Jul/24

$i f t h e s e r i e s \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} c o n v e r g e s t o k . f i n d t h e c o n v e r g e n c e v a l u e o f \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n + 1)}^{2}}$
	Answered by mr W last updated on 23/Jul/24

	$\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} = k$ $\underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n)}^{2}} = \frac{k}{4}$ $\underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n)}^{2}} + \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n + 1)}^{2}} + \frac{1}{1^{2}} = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} = k$ $\frac{3 k}{4} + \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n + 1)}^{2}} + 1 = k$ $\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n + 1)}^{2}} = \frac{3 k}{4} - 1 ✓$
	Answered by lmcp1203 last updated on 23/Jul/24

	$t h a n k s$
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