# Does $\sum_{n=0}^\infty \frac{3n}{n^2+2n}$ converge?

My attempts:

Ratio-test: Inconclusive

Comparision-test (to harmonic/ geometric series):

$$\sum_{n=0}^\infty \frac{3n}{n^2+2n} = \sum_{n=0}^\infty \frac{3}{n+2} = 3 \cdot\sum_{n=0}^\infty \frac{1}{n+2}$$

The series is a little bit smaller than the harmonic series $$\sum_{n=0}^\infty \frac{1}{n}$$ and im therefore unsure.

• It's the harmonic series with one term removed. If you remove a finite number from infinity, what do you get? Commented Sep 1, 2017 at 19:34
• Use limit-comparison test Commented Sep 1, 2017 at 20:11
• It's smaller by a finite number. It smaller by 1. (BTW: harmonic series starts at $1$; not $0$). Commented Sep 1, 2017 at 20:24

HINT: use that $$\frac{1}{n+2}\geq \frac{1}{2n}$$

Hint: Note that $$\frac{3n}{n^2 + 2n} \geq \frac{3n}{n^2 + 2n^2} = \frac 1n$$

Compare it to $\sum_{n=2}^\infty\frac1n$ instead. Or note that the $3\cdot{}$ makes it larger than the harmonic series again.

The shortest way is, as often, with equivalents: $\;n^2+2n\sim_\infty n^2$, so $$\frac{3n}{n^2+2n}\sim_\infty\frac{3n}{n^2}=\frac3n,\quad\text{which diverges.}$$

Note: you can reindex:

$\sum_{n=0}^{\infty} \frac 1{n+2} = \sum_{m=2}^{\infty} \frac 1m = \sum_{m=1}\frac 1m -1$.

Which we can use to show $\sum_{n= 0}^{\infty} \frac 1{n+k}$ will always diverge no matter how large a $k$ is.

$\sum\limits_{j=1}^{k-1}\frac 1j + \sum_\limits{n= 0}^{\infty} \frac 1{n+k}= \sum_\limits{j= 1}^{\infty}\frac 1j$. We know $\sum\limits_{j=1}^{k-1}\frac 1j$ is a finite value, so $\sum_\limits{n= 0}^{\infty} \frac 1{n+k}$ converges if and only if $\sum_\limits{j= 1}^{\infty}\frac 1j$ does. And we know it doesn't.

So you sum would only converge if the harmonic series coverged (and it would be $3(N - 1)$ if $N$ were the sum of the harmonic series). But the harmonic series doesn't converge.

You can apply the Integral Test. Let $f(x)=\frac{3x}{x^2+x}$ be the function describing the series' n-th term. You can conclude that $f(x) > 0$ is continuous for every $x \in [1, +oo[$. You can derive the function and conclude that $\frac{d}{dx}f(x) < 0$ for the given interval. Notice that you have all the conditions satisfied to apply the test. Integrating $\int_{1}^{+oo}$ will lead you to conclude that the series diverges.