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Question Number 202468 by hardmath last updated on 27/Dec/23

$$\mathrm{Find}:$$$$1.\underset{\mathbf{n}=1}{\sum ^{\mathrm{\infty }}}\frac{16}{{16\mathrm{n}}^2-8\mathrm{n}-3}=?2.\underset{\mathbf{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathbf{n}}}{{2\mathrm{n}}^3}=?$$

Answered by Rasheed.Sindhi last updated on 27/Dec/23

$$1.\underset{n=1}{\stackrel{\mathrm{\infty }}{\mathrm{\Sigma }}}\frac{16}{16n^2-8n-3}$$$$\frac{16}{16n^2-8n-3}=\frac{16}{(4n-3)(4n+1)}=\frac{a}{4n-3}+\frac{b}{4n+1}$$$$16=a(4n+1)+b(4n-3)$$$$n=-1/4:16=-4b\Rightarrow b=-4$$$$n=\frac{3}{4}:16=4a\Rightarrow a=4$$$$\underset{n=1}{\stackrel{\mathrm{\infty }}{\mathrm{\Sigma }}}\frac{16}{16n^2-8n-3}=\underset{n=1}{\stackrel{\mathrm{\infty }}{\mathrm{\Sigma }}}(\frac{4}{4n-3}-\frac{4}{4n+1})$$$$\begin{array}{|cc|}\hline t_1& \frac{4}{1}-\cancel{\frac{4}{5}}\\ t_2& \cancel{\frac{4}{5}}-\cancel{\frac{4}{9}}\\ t_3& \cancel{\frac{4}{9}}-\cancel{\frac{4}{13}}\\ ...& ...\\ t_{n-1}& \cancel{\frac{4}{4n-7}}-\cancel{\frac{4}{4n-3}}\\ t_n& \cancel{\frac{4}{4n-3}}-\frac{4}{4n+1}\\ & \underset{n=1}{\sum ^{n=n}}\frac{4}{4n-3}-\frac{4}{4n+1}=4-\frac{4}{4n+1}\\ \hline\end{array}$$$$\underset{n=1}{\stackrel{\mathrm{\infty }}{\mathrm{\Sigma }}}(4-\frac{4}{4n+1})=\underset{x\to \mathrm{\infty }}{\lim }(4-\frac{4}{4n+1})=4-0=4$$

Commented by hardmath last updated on 30/Dec/23

$$\mathrm{thankyou}\mathrm{dear}\mathrm{professor}\mathrm{cool}$$

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