Prove whether the following series is convergent or divergent

Question

Prove the convergence of the following series: $$\sum\frac{1 + 2^n + 5^n}{3^n}$$ My idea was to write it as: $$\frac{1}{3^n} + \frac{2^n}{3^n} + \frac{5^n}{3^n}$$ Which would mean that $\frac{1}{3^n}$ is a geometric series with a ratio of $\frac{1}{3}$, and since $\frac{1}{3} <1$,$\frac{1}{3^n}$ is convergent. However, I don't know what to do with the rest...

Answer 1

There is a hazard in the manipulation you have performed. $$ \lim_{N \rightarrow \infty} \left( f(N) + g(N) \right) = \lim_{N \rightarrow \infty} f(N) + \lim_{N \rightarrow \infty} g(N) $$ only if the two limits on the right exist. This condition prevents the absurdity $$ \lim_{N \rightarrow \infty} \sum_{i=1}^N 0 \overset{?}= \lim_{N \rightarrow \infty} \sum_{i=1}^N 1 + \lim_{N \rightarrow \infty} \sum_{i=1}^N -1 \text{.} $$

So, you method works if all three limits that you obtain exist, but the third one does not. This suggests that the original sum might not converge, and/or a different technique will have to be used to determine convergence/divergence.

Notice that for $n \geq 0$, $$ 1 + 3^n + 5^n > 2 + 5^n > 5^n \text{.} $$ So $$ \sum_{n=1}^N \frac{1 + 2^n + 5^n}{3^n} > \sum_{n=1}^N \frac{5^n}{3^n} = \sum_{n=1}^N \left( \frac{5}{3} \right)^n \text{,} $$ a geometric series whose convergence or divergence you should already be able to determine.

If the partial sums of positive terms is always greater than the same (meaning: same upper bound on the index) partial sums of a series that diverges to $+\infty$, what can you say about the first sum?

Likewise, if the partial sums of positive terms is always less than the same partial sums of a series that converges, what can you say about the first sum?

Answer 2

Simply write $\frac{5^n}{3^n}$ as $\left(\frac53\right)^n$. Then, $$\sum \frac{1+2^n+5^n}{3^n}>\sum \frac{5^n}{3^n}=\sum\left(\frac53\right)^n>\sum 1^n=\infty$$ So the entire sum diverges.