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calculate

$\lim_{n\to\infty}\frac{n}{a^{n+1}}$( $a$+$\frac {a^2}{2} +.....+\frac{a^n}{n}$), where $a >1$.

here i take $x_n$ = $a$ + $\frac {a^2}{2} +.....+\frac{a^n}{n}$ and $y_n$ =$\frac {a^{n+1}}{n}$

now $\lim_{n\to\infty}$ =$\frac {x_n }{y_n}$ =$\frac{n}{a^{n+1}}a$ + $\frac{n}{a^{n+1}}\frac {a^2}{2}$ +...........+$\frac{n}{a^{n+1}}$$\frac{a^n}{n}$ = $\infty$

Is its corrects ????

Pliz help me

thanks in advance

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    Hint: Can you justify (and use) the limit $$\sum_{k=1}^n\frac{n}{n-k+1}\frac1{a^k}\longrightarrow\sum_{k=1}^\infty\frac1{a^k}\ ?$$ – Did Mar 25 '18 at 07:19
  • im not getting @Did –  Mar 25 '18 at 07:21
  • That may be so, but basically waiting for a full solution to pop up is not going to lead you very far, with respect to your mathematical understanding. As they say, mathematics is not a spectator's sport. – Did Mar 25 '18 at 07:38

1 Answers1

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HINT

Let consider

$$\frac{a_n}{b_n}=\frac{\sum_{k=1}^n \frac{a^k}{k}} {\frac{a^{n+1}}{n}}$$

and apply Stolz-Cesaro

$$\lim_{n \to \infty} \frac{a_n}{b_n}=\lim_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$

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  • thanks a lots @gimusi,,,,,keep us helping always ,,as i got the answer $\frac{1}{a-1}$ –  Mar 25 '18 at 10:40