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.
$\begingroup$
$\endgroup$
1
-
$\begingroup$ $\sum_{n=1}^N\left(\frac{4n+3}{5^n}\right)=2-\frac{2+N}{5^N}$ $\endgroup$– user140541Jun 21, 2015 at 1:29
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
3
$$\sum_{n=0}^{\infty }\frac{4n+3}{5^n}=\sum_{n=0}^{\infty }\frac{4n}{5^n}+\sum_{n=0}^{\infty }\frac{3}{5^n}$$ $$\sum_{n=0}^{\infty }x^n=\frac{1}{1-x}$$ 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 $x=1/5$
-
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 $1$ $\endgroup$– user249369Jun 21, 2015 at 1:45
-
$\begingroup$
$\endgroup$
Hint: $$\sum_{n=0}^\infty \frac{n}{5^n}=\sum_{n=1}^\infty\sum_{m=n}^\infty \frac{1}{5^m}$$
$\begingroup$
$\endgroup$
1
a telescoping way,
$\sum_{n=0}^\infty\left(\frac{n+1}{5^{n-1}}\right)$$-$ $\sum_{n=0}^\infty\left(\frac{n+2}{5^n}\right)$
-
$\begingroup$ I was thinking in this line. But we need to prove the convergence, right? Also, how to derive this relation? $\endgroup$ Jun 21, 2015 at 2:52
$\begingroup$
$\endgroup$
$$ \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 $\sum_{n=0}^{\infty}\frac{4n+3}{5^n}=5$.
