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Let $S_{k}$ is defined as $\sum^{\infty}_{n=1}\bigg(\frac{1}{k}\bigg)^n$ where $k$ is positive integer $>1$. Then minimum and maximum value of $\displaystyle \frac{S_{k}S_{k+2}}{(S_{k+1})^2}$

Try: $$\sum^{\infty}_{n=1}\bigg(\frac{1}{k}\bigg)^n=\frac{1}{k-1}$$

So $$\displaystyle \frac{S_{k}S_{k+2}}{(S_{k+1})^2}=\frac{k^2}{k^2-1}$$ Then minimum value of $$f(k)=1$$ when $k{\rightarrow \infty}$.

Could some help me how to find maximum value, Thanks

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Let $f(x):= \frac{x^2}{x^2-1}$ for $x \ge 2$. Then $f'(x)<0$, hence $f$ is decreasing. Thus

$\max \{f(x): x \ge 2\}=f(2) = \frac{4}{3}$.

$\min \{f(x): x \ge 2\}$ does not exist, but $\inf \{f(x): x \ge 2\}=1$.

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Rewrite it as $$\frac{k^2}{k^2-1}=1+\frac 1{k^2-1}$$ so the maximum value is attained at the minimum value of $k$, i.e. the maximum value is $\:1+\dfrac1{4-1}=\dfrac43$.

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  • $\begingroup$ No. The maximum value is $4/3$. $\endgroup$ – Fred Mar 27 '18 at 9:43
  • $\begingroup$ @Fred: Thanks for pointing the slip! Fixed. $\endgroup$ – Bernard Mar 27 '18 at 9:46

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