Let $$\sum {n\over{n+1}} \cdot \left({{2x+1} \over x}\right)^n$$
I was asked to calculate the radius of convergence. We can write the series as:
$$\sum {n\over {n+1}}\cdot \left(2+{1\over x}\right)^n$$
Now, we can define $t:={2+{1\over x}}$ and then evaluate the series $\displaystyle\sum {n\over {n+1}}\cdot t^n$.
Is there another way I'm expected to think of without using this "trick"?
Take root test, $$\lim_{n\to\infty} \sqrt [n] { \left| \frac{n}{n+1} \cdot \left( \frac{2x+1}{x}\right)^n \right |} < 1$$ $$ \mathrm{or, }\frac{1}{\lim_{n\to\infty}\left( 1 + \frac 1 n \right)^{\frac 1 n}} \left| \frac{2x+1}{x}\right| < 1 $$