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Question Number 92003 by zainal tanjung last updated on 04/May/20

$$\mathrm{The}\mathrm{sum}\mathrm{to}n\mathrm{terms}\mathrm{of}\mathrm{the}\mathrm{series}$$$$\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+....\mathrm{is}$$

Commented by Prithwish Sen 1 last updated on 04/May/20

$${\mathrm{t}}_{\mathrm{n}}=\frac{6(2\mathrm{n}+1)}{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}=6[\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}]$$$$\mathrm{now}\mathrm{the}\mathrm{series}\mathrm{becomes}$$$$\underset{\mathrm{k}\to \mathrm{\infty }}{\lim }6\left[\underset{\mathrm{n}=1}{\sum ^{\mathrm{k}}}(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1})\right]=\underset{\mathrm{k}\to \mathrm{\infty }}{\lim }6\left[\frac{\mathrm{k}}{\mathrm{k}+1}\right]=$$$$\underset{\mathrm{k}\to \mathrm{\infty }}{\lim }6.\frac{1}{1+\frac{1}{\mathrm{k}}}=6.1=6\mathbf{please}\mathbf{check}$$

Commented by mathmax by abdo last updated on 04/May/20

$$S=\sum _{n=1}^{\mathrm{\infty }}\frac{2n+1}{\sum _{p=1}^n{p}^2}\Rightarrow S=\sum _{n=1}^{\mathrm{\infty }}\frac{2n+1}{\frac{n(n+1)(2n+1)}{6}}$$$$=6\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)}=6{lim}_{n\to +\mathrm{\infty }}\sum _{k=1}^n\frac{1}{k(k+1)}$$$$=6{lim}_{n\to +\mathrm{\infty }}\sum _{k=1}^n(\frac{1}{k}-\frac{1}{k+1})$$$$=6{lim}_{n\to +\mathrm{\infty }}\{1-\frac{1}{n+1}\}=6$$

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