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

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

$$ \\ $$$${if}\:{the}\:{series}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:{converges}\:{to}\:{k}\:.\:\:{find}\:\:{the}\:{convergence}\:{value}\:{of}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Answered by mr W last updated on 23/Jul/24

Σ_(n=1) ^∞ (1/n^2 )=k Σ_(n=1) ^∞ (1/((2n)^2 ))=(k/4) Σ_(n=1) ^∞ (1/((2n)^2 ))+Σ_(n=1) ^∞ (1/((2n+1)^2 ))+(1/1^2 )=Σ_(n=1) ^∞ (1/n^2 )=k ((3k)/4)+Σ_(n=1) ^∞ (1/((2n+1)^2 ))+1=k ⇒Σ_(n=1) ^∞ (1/((2n+1)^2 ))=((3k)/4)−1 ✓

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }={k} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{2}} }=\frac{{k}}{\mathrm{4}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{2}} }+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }={k} \\ $$$$\frac{\mathrm{3}{k}}{\mathrm{4}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{1}={k} \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{3}{k}}{\mathrm{4}}−\mathrm{1}\:\checkmark \\ $$

Answered by lmcp1203 last updated on 23/Jul/24

thanks

$${thanks} \\ $$

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