# Thread: power series help

1. ## power series help

My professor says that this will can be evaluated using the integral test, but i can't figure out what to do: please help! i have my final tomorrow and though this is the last problem, i don't understand it.

∑(n=2 to infinity) of (2^n*(x+1)^n)/(n*ln(n))

2. ## Re: power series help

Originally Posted by jaqueh
My professor says that this will can be evaluated using the integral test, but i can't figure out what to do: please help! i have my final tomorrow and though this is the last problem, i don't understand it.

∑(n=2 to infinity) of (2^n*(x+1)^n)/(n*ln(n))
Have you tried the ratio test? The interval is convergent where \displaystyle \displaystyle \begin{align*} \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right| < 1 \end{align*}, so

\displaystyle \displaystyle \begin{align*} \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| &<1 \\ \lim_{n \to \infty}\left|\frac{\frac{2^{n + 1}(x + 1)^{n + 1}}{(n + 1)\ln{(n + 1)}}}{\frac{2^n(x + 1)^n}{n\ln{n}}}\right| &<1 \\ \lim_{n \to \infty}\left|\frac{2^{n + 1}(x + 1)^{n + 1}n\ln{n}}{2^n(x + 1)^n(n + 1)\ln{(n + 1)}}\right| &< 1 \\ \lim_{n \to \infty}\left|\frac{2(x + 1)n\ln{n}}{(n + 1)\ln{(n + 1)}}\right| &< 1 \\ \lim_{n \to \infty}\left|\frac{2n\ln{n}}{(n + 1)\ln{(n+1)}}\right||x + 1| &< 1 \\ \lim_{n \to \infty}2\left|\frac{\ln{n} + 1}{\ln{(n+1)} + 1}\right||x + 1| &< 1 \textrm{ by L'Hospital's Rule} \\ \lim_{n \to \infty}2\left|\frac{\frac{1}{n}}{\frac{1}{n + 1}}\right||x + 1| &< 1 \textrm{ by L'Hospital's Rule again...} \\ \lim_{n \to \infty}2\left|\frac{n + 1}{n}\right||x + 1| &< 1 \\ \lim_{n \to \infty}2\left|1 + \frac{1}{n}\right||x + 1| &< 1 \\ 2(1)|x + 1| &< 1 \\ |x + 1| &< \frac{1}{2} \\ -\frac{1}{2} < x + 1 &< \frac{1}{2} \\ -\frac{3}{2} < x &< -\frac{1}{2} \end{align*}

You will also need to check endpoints.

3. ## Re: power series help

Thank you! that answered my question!