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Question Number 88000 by jdmath last updated on 07/Apr/20

Is there a formula to calculate     Σ_(i=1) ^n (1/i^2 )  interms of n..?

$${Is}\:{there}\:{a}\:{formula}\:{to}\:{calculate}\: \\ $$$$ \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{i}^{\mathrm{2}} } \\ $$$${interms}\:{of}\:{n}..? \\ $$

Commented by abdomathmax last updated on 08/Apr/20

i thnk?no?formulae but we can write  Σ_(i=1) ^(n )  (1/i^2 ) =Σ_(i=3k)  (...)+Σ_(i=3k+1)  (...)+Σ_(i=3k+2)   (...)  =Σ_(k=1) ^([(n/3)])  (1/(9k^2 )) +Σ_(k=0) ^([((n−1)/3)])  (1/((3k+1)^2 )) +Σ_(k=0) ^([((n−2)/3)])  (1/((3k+2)^2 ))  =....

$${i}\:{thnk}?{no}?{formulae}\:{but}\:{we}\:{can}\:{write} \\ $$$$\sum_{{i}=\mathrm{1}} ^{{n}\:} \:\frac{\mathrm{1}}{{i}^{\mathrm{2}} }\:=\sum_{{i}=\mathrm{3}{k}} \:\left(...\right)+\sum_{{i}=\mathrm{3}{k}+\mathrm{1}} \:\left(...\right)+\sum_{{i}=\mathrm{3}{k}+\mathrm{2}} \:\:\left(...\right) \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\left[\frac{{n}}{\mathrm{3}}\right]} \:\frac{\mathrm{1}}{\mathrm{9}{k}^{\mathrm{2}} }\:+\sum_{{k}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{3}}\right]} \:\frac{\mathrm{1}}{\left(\mathrm{3}{k}+\mathrm{1}\right)^{\mathrm{2}} }\:+\sum_{{k}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{2}}{\mathrm{3}}\right]} \:\frac{\mathrm{1}}{\left(\mathrm{3}{k}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$=.... \\ $$

Commented by jdmath last updated on 16/Apr/20

thank you...rly

$${thank}\:{you}...{rly} \\ $$

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