# does the series converge? $\sum_{n=1}^\infty\left(\frac {3}{5^n}+\frac 2n\right)$

does the series converge? $$\sum_{n=1}^\infty\left(\frac {3}{5^n}+\frac 2n\right)$$

now the $\frac 2n$ should diverge (Harmonic series), but I keep getting stuck trying to evaluate the first sum. By the comparison test they shouldn't both diverge but I can't figure out how to get the answer.

• Hint: the first sum is a geometric series – Mathmo123 Jun 30 '14 at 20:48
• If two series converge then it sum converges. Can you apply this? – mfl Jun 30 '14 at 20:48
• @mfl The second series doesn't converge. Convergent series + divergent series = divergent series. – becko Jun 30 '14 at 20:53
• @becko I was thinking of arguing by contradiction. If this series were convergent then its sum with (minus) the geometric series would be convergent, which is not true. – mfl Jun 30 '14 at 20:55

You just need to say that

$$\frac 3 {5^n} + \frac 2 n \geq \frac 2 n$$

and apply the comparison test.

• so my $a_n$ in this case is to the left of the sign? I was treating the $\frac{3}{5^n}$ as the $a_n$ which looks to be incorrect. – Joshhw Jun 30 '14 at 20:55
• @Joshhw I'm not sure what you mean by $a_n$. Each term of the series in your question is greater than the corresponding term in the harmonic series. If the terms in one (positive) series are greater than the terms in a divergent (positive) series, then of course the former series diverges too. – Jack M Jun 30 '14 at 21:06
• I'm not sure how to determine what is $a_n$ and what to consider $b_n$. I was treating the equation like $\frac {3}{5^n}$ as the $a_n$ in the comparison test and $\frac 2n$ as the $b_n$ – Joshhw Jun 30 '14 at 21:09
• @Joshhw It's possible we're not talking about the same comparison test. I'm saying that if $\sum b_n$ converges and $0\leq a_n\leq b_n$ for all $n$, then $\sum a_n$ converges. – Jack M Jun 30 '14 at 21:12

Another way to do it: Assume that the series is convergent. Note that $$\sum \frac{3}{5^n}$$ is convergent. The difference of two convergent series is convergent. So $$\sum \frac{3}{5^n} + \frac{2}{n} - \sum \frac{3}{5^n} = \sum \frac{2}{n}$$ would be convergent (contradiction).

The first term converges since it is just a geometric series. However a constant added to a divergent series doesn't change the convergence behavior of the series: the series still diverges by the comparison test.