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Question Number 174361 by infinityaction last updated on 30/Jul/22

Commented by a.lgnaoui last updated on 31/Jul/22

$$S=\frac{5\times 3}{(5-3)(5^2-3^2)}+\frac{5^2\times 3^2}{(5^2-3^2)(5^3-3^3)}+\frac{5^3\times 3^3}{(5^3-3^3)(5^4-3^4)}+....+\frac{5^n\times 3^n}{(5^n-3^n)(5^{n+1}-3^{n+1})}$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{5^n{3}^n}{(5^n-3^n)(5^{n+1}-3^{n+1})}$$$$\frac{5^n{3}^n}{(5^n-3^n)(5^{n+1}-3^{n+1})}=\frac{1}{(\frac{1}{3^n}-\frac{1}{5^n})(\frac{1}{3^{n+1}}-\frac{1}{5^{n+1}})}$$$$S=\sum _{n=1}^{\mathrm{\infty }}\left[\frac{1}{(\frac{1}{3^n}-\frac{1}{5^n})(\frac{1}{3^{n+1}}-\frac{1}{5^{n+1}})}\right]$$$$........$$

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