Ratio test results argue Wolfram

Question

$$\sum\limits_{n=1}^{\infty} \frac{1+2^n+3^n}{5^n}$$

Going through with the ratio test, I eneded up with $ L = 1 $ which means the test is inconclusive. Wolfram, however, says that the series converges by ratio test

Answer 1

Combine the ratio test with equivalence: $\;1+2^n+3^n\sim_\infty 3^n$, hence $$\frac{1+2^n+3^n}{5^n}\sim_\infty \frac{3^n}{5^n}=\Bigl(\frac35\Bigr)^n,$$ which converge.

Answer 2

The ratio test asks us to find $L=\lim\limits_{n\to\infty}|\frac{a_{n+1}}{a_n}|$ where $a_n$ is the $n^\text{th}$ summand in our series. Noting that all terms are positive, I drop the use of the absolute value sign here.

$\lim\limits_{n\to\infty}\dfrac{\left(\frac{1+2^{n+1}+3^{n+1}}{5^{n+1}}\right)}{\left(\frac{1+2^n+3^n}{5^n}\right)}$

$=\lim\limits_{n\to\infty}\frac{1+2^{n+1}+3^{n+1}}{(1+2^n+3^n)5}$

$=\lim\limits_{n\to\infty}\frac{1+2^{n+1}+3^{n+1}}{(1+2^n+3^n)5}\cdot \frac{3^{-n-1}}{3^{-n-1}}$

$=\lim\limits_{n\to\infty}\frac{(\frac{1}{3})^{n+1}+(\frac{2}{3})^{n+1}+1}{((\frac{1}{3})^{n+1}+\frac{1}{3}(\frac{2}{3})^{n}+\frac{1}{3})5}$

All terms in the above expression go to zero except for the constants.

$=\frac{3}{5}$

As the limit approaches a number between zero and one, the ratio test tells us that the series is indeed convergent.

Remember that $\frac{x^{n+1}+y^{n+1}+z^{n+1}}{x^n+y^n+z^n}\neq x+y+z$. That might have been the source of your confusion.