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Question Number 39517 by math khazana by abdo last updated on 07/Jul/18

find radius of S(x)=Σ_(n=1) ^∞ (x^n /n^2 ) and calculate its sum 2) find Σ_(n=1) ^∞ (1/n^2 ) and Σ_(n=1) ^∞ (1/(n^2 2^n )) .

$${find}\:{radius}\:{of}\:\:{S}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}^{\mathrm{2}} } \\ $$$${and}\:{calculate}\:{its}\:{sum} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\:\:{and}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:\mathrm{2}^{{n}} }\:. \\ $$$$ \\ $$

Commented by abdo mathsup 649 cc last updated on 07/Jul/18

for x≠0 ∣ ((u_(n+1) (x))/(u_n (x)))∣ =∣ ((x^(n+1) /((n+1)^2 ))/(x^n /n^2 ))∣=(n^2 /((n+1)^2 )) ∣x∣→∣x∣ so if ∣x∣<1 the serie converges if x=−1 Σ (((−1)^n )/n^2 ) converges(alternate serie) if x=1 Σ (1/n^2 ) is a reiman serie convergent and R=1 we have (dS/dx)(x)= Σ_(n=1) ^∞ ((nx^(n−1) )/n^2 ) =(1/x)Σ_(n=1) ^∞ (x^n /n) =−((ln(1−x))/x) ( we suppose that x≠0) ⇒ S(x) = −∫_0 ^x ((ln(1−t))/t) dt +c c=S(0)=0 ⇒ S(x)=−∫_0 ^x ((ln(1−t))/t) dt .

$${for}\:{x}\neq\mathrm{0}\:\:\mid\:\frac{{u}_{{n}+\mathrm{1}} \left({x}\right)}{{u}_{{n}} \left({x}\right)}\mid\:=\mid\:\frac{\frac{{x}^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }}{\frac{{x}^{{n}} }{{n}^{\mathrm{2}} }}\mid=\frac{{n}^{\mathrm{2}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\mid{x}\mid\rightarrow\mid{x}\mid \\ $$$${so}\:{if}\:\mid{x}\mid<\mathrm{1}\:\:{the}\:{serie}\:{converges} \\ $$$${if}\:{x}=−\mathrm{1}\:\:\Sigma\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:{converges}\left({alternate}\:{serie}\right) \\ $$$${if}\:{x}=\mathrm{1}\:\:\Sigma\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\:{is}\:{a}\:{reiman}\:{serie}\:{convergent} \\ $$$${and}\:{R}=\mathrm{1} \\ $$$${we}\:{have}\:\:\frac{{dS}}{{dx}}\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{nx}^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{x}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}} \\ $$$$=−\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}\:\:\left(\:\:{we}\:{suppose}\:{that}\:{x}\neq\mathrm{0}\right)\:\Rightarrow \\ $$$${S}\left({x}\right)\:=\:−\int_{\mathrm{0}} ^{{x}} \:\:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}\:{dt}\:+{c} \\ $$$${c}={S}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow\:{S}\left({x}\right)=−\int_{\mathrm{0}} ^{{x}} \:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}\:{dt}\:. \\ $$

Commented by abdo mathsup 649 cc last updated on 07/Jul/18

2) Σ_(n=1) ^∞ (1/n^2 ) =S(1) = −∫_0 ^1 ((ln(1−t))/t) dt and we have always proved that Σ_(n=1) ^∞ (1/n^2 ) =(π^2 /6) .

$$\left.\mathrm{2}\right)\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:={S}\left(\mathrm{1}\right)\:\:=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}\:\:{dt}\: \\ $$$${and}\:{we}\:{have}\:{always}\:{proved}\:{that}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:. \\ $$

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