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Question Number 60503 by prof Abdo imad last updated on 21/May/19
$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}+\mathrm{2}+\mathrm{3}+...+{n}}{\mathrm{1}^{\mathrm{3}} \:+\mathrm{2}^{\mathrm{3}} \:+\mathrm{3}^{\mathrm{3}} \:+...+{n}^{\mathrm{3}} } \\ $$
Answered by Prithwish sen last updated on 21/May/19
![t_n =(((n(n+1))/2)/([((n(n+1))/2)]^2 )) =2[(1/n)−(1/(n+1))] ∴S_n =2[1−(1/(n+1))] Now as n→∞ ΣS_n →2 waiting for your feedback.](Q60515.png)
$$\mathrm{t}_{\mathrm{n}} =\frac{\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}}{\left[\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} } \\ $$$$\:=\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right] \\ $$$$\therefore\mathrm{S}_{\mathrm{n}} =\mathrm{2}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right] \\ $$$$\mathrm{Now}\:\mathrm{as}\:\mathrm{n}\rightarrow\infty \\ $$$$\Sigma\mathrm{S}_{\mathrm{n}} \rightarrow\mathrm{2} \\ $$$$\mathrm{waiting}\:\mathrm{for}\:\mathrm{your}\:\mathrm{feedback}. \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 21/May/19
$${correct}\:{sir}\:{thanks} \\ $$