Prove whether the following series is convergent or divergent 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...
 A: 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.
A: 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?
