if-the-series-n-1-1-n-2-converges-to-k-find-the-convergence-value-of-n-1-1-2n-1-2-

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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