find exact values for the summation

[spdrmncoo]

find exact values for the summation... means to find x.
but where and how can i change it

sigma [ n= 1 to infinity] n/ 4^ (n+1)

to the exspression with x and after to find x? :?:

 

[galactus]

Did you look at the other I answered.

\(\displaystyle \L\\\sum_{n=1}^{\infty}\frac{n}{a^{n+1}}=\frac{1}{a}\sum_{n=1}^{\infty}\frac{n}{a^{n}}=\frac{1}{(a-1)^{2}}\)

 

[spdrmncoo]

thank you

 

[soroban]

Hello, spdrmncoo!

Another approach . . .

Find exact value of: \(\displaystyle \L\:\sum^{\infty}_{n=1}\frac{n}{4^{n+1}}\)

Expand the summation: \(\displaystyle \L\:S\;=\;\frac{1}{4^2}\,+\,\frac{2}{4^3}\,+\,\frac{3}{4^4}\,+\,\frac{4}{4^5}\,+\,\frac{5}{4^6}\,+\,\cdots\)

. . . . . . Multiply by \(\displaystyle \frac{1}{4}:\L\;\;\frac{1}{4}S \;=\;\qquad\qquad\qquad\quad\frac{1}{4^3}\,+\,\frac{2}{4^4}\,+\,\frac{3}{4^5}\,+\,\frac{4}{4^6}\,+\,\cdots\)


Subtract: \(\displaystyle \L\:\frac{3}{4}S\;=\;\frac{1}{4^2} \,+\,\frac{1}{4^3}\,+\,\frac{1}{4^4}\,+\,\frac{1}{4^5}\,+\,\frac{1}{4^6}\,+\,\cdots\)


The right side is a geometric series with first term $\displaystyle a\,=\,\frac{1}{16}$ and common ratio $\displaystyle r\,=\,\frac{1}{4}$
. . Its sum is: $\displaystyle \:\frac{\frac{1}{16}}{1\,-\,\frac{1}{4}} \:=\:\frac{1}{12}$


Therefore: \(\displaystyle \L\:\frac{3}{4}S\,=\,\frac{1}{12}\;\;\Rightarrow\;\;\fbox{S\,=\,\frac{1}{9}}\)