Question

Write the following power series as a function

d) \(\sum_{n=1}^{\infty}\frac{1}{n}x^{n+3}\)

Answer 1

\[\sum_{n=1}^{\infty}\frac{1}{n}x^{n+3}=x^3\sum_{n=1}^{\infty}\frac{1}{n}x^{n}=x^3\sum_{n=1}^{\infty}\left (\int x^{n-1}\ dx\right )\]

Answer 2

is it end solution? or it needs to be add something more?

Answer 3

No, it's not the end yet.
\[x^3\sum_{n=1}^{\infty}\left (\int x^{n-1}\ dx\right )=x^3 \int \left ( \sum_{n=1}^\infty x^{n-1}\right )dx\]
Can you see what to do now?

Answer 4

no not really, but you can go on with your solution, i try to figure it out

Answer 5

\[\sum_{n=1}^\infty x^{n-1}= \frac{1}{x^2}\sum_{n=0}^\infty x^n=\frac{1}{x^2(1-x)}\]
Now integrate this, and multiply by x^3 and that's the answer.

Answer 6

ok i will try but i am not sure if can if you can bring it to end it will be much appreciate, i am noob at mathematics and must give homework soon

Answer 7

:(

Answer 8

\[\frac{1}{x^2(1-x)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{1-x}\\
1=A(x)(1-x)+B(1-x)+Cx^2\\
1=Ax-Ax^2+B-Bx+Cx^2\\
1=(C-A)x^2+(A-B)x+B \\
A=B=C=1\\
\int \frac{1}{x^2(1-x)}\ dx= \int \frac{1}{x}+\frac{1}{x^2}+\frac{1}{1-x}\ dx\\
=\ln\left (\frac{x}{1-x} \right )-\frac{1}{x}\\
\sum_{n=1}^{\infty}\frac{1}{n}x^{n+3}=x^3 \ln\left (\frac{x}{1-x} \right )-x^2\]

Answer 9

now this is end solutioin right?

Answer 10

It's the end.

Answer 11

ok thank you very very much