0
$\begingroup$

What is the sum of the series $\sum_{n=0}^\infty\left(\frac{4n+3}{5^n}\right)$ ? I got that the series converges and the sum seems to be $5$. When trying to explicitly get the sum, I tried to find the appropriate function whose Maclaurin series would be this sum.

$\endgroup$
1
  • $\begingroup$ $\sum_{n=1}^N\left(\frac{4n+3}{5^n}\right)=2-\frac{2+N}{5^N}$ $\endgroup$
    – d.k.o.
    Jun 21, 2015 at 1:29

4 Answers 4

4
$\begingroup$

$$\sum_{n=0}^{\infty }\frac{4n+3}{5^n}=\sum_{n=0}^{\infty }\frac{4n}{5^n}+\sum_{n=0}^{\infty }\frac{3}{5^n}$$ $$\sum_{n=0}^{\infty }x^n=\frac{1}{1-x}$$ derive it $$\sum_{n=1}^{\infty }nx^{n-1}=\frac{1}{(1-x)^2}$$ multiply by $x$ $$\sum_{n=1}^{\infty }nx^{n}=\frac{x}{(1-x)^2}$$ at $x=1/5$ $$\sum_{n=1}^{\infty }\frac{4n}{5^{n}}=4\frac{1/5}{(1-1/5)^2}$$ the second series is geometric series $$3\sum_{n=0}^{\infty }x^n=3\left(\frac{1}{1-x}\right)$$ find the value at $x=1/5$

$\endgroup$
3
  • 1
    $\begingroup$ You really don't need to subtract one before taking the derivative. $\endgroup$ Jun 21, 2015 at 1:39
  • $\begingroup$ @ThomasAndrews but I think the $n$ was $1$ $\endgroup$
    – user249369
    Jun 21, 2015 at 1:45
  • $\begingroup$ Yes, but what is $nx^n$ when $n=0$? $\endgroup$ Jun 21, 2015 at 1:46
2
$\begingroup$

Hint: $$\sum_{n=0}^\infty \frac{n}{5^n}=\sum_{n=1}^\infty\sum_{m=n}^\infty \frac{1}{5^m}$$

$\endgroup$
2
$\begingroup$

a telescoping way,

$\sum_{n=0}^\infty\left(\frac{n+1}{5^{n-1}}\right)$$-$ $\sum_{n=0}^\infty\left(\frac{n+2}{5^n}\right)$

$\endgroup$
1
  • $\begingroup$ I was thinking in this line. But we need to prove the convergence, right? Also, how to derive this relation? $\endgroup$ Jun 21, 2015 at 2:52
1
$\begingroup$

$$ \begin{align} \sum_{n=0}^{\infty}(4n+3)x^n&=4\sum_{n=0}^{\infty}(n+1)x^n\sum_{n=0}^{\infty}x^n\\ &=4(\sum_{n=0}^{\infty}x^{n+1})'-\sum_{n=0}^{\infty}x^n\\ &=4(\frac{x}{1-x})'-\frac{1}{1-x}\\ &=\frac{x+3}{(1-x)^2} \end{align} $$

Set $x=\frac{1}{5}$, we have $\sum_{n=0}^{\infty}\frac{4n+3}{5^n}=5$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.