Determine whether the following are convergent or divergent. If it is convergent find its sum. Make sure to fully justify all of your work.

$$(a) \sum_{n=2}^\infty \frac{3}{n^2+3n}$$

Solution (My attempt):

Let $f(n) = a_n$

On $[2,\infty), f(x) = \frac 3 {x^2+3x} > 0$

$$f'(x) = 3\left(\frac{-(2x+3)}{(x^2+3x)^2}\right) < 0, \forall x \in [2,\infty)$$

Therefore the hypothesis of the integral test is met.


$$\lim_{A\to\infty} \int_2^A \frac 3 {(x)(x+3)} \, dx$$

$$\lim_{A\to\infty} \int_{2}^{A} \left( \frac{1}{x} - \frac{1}{x+3} \right)dx \text{ By pfd}$$

$$\lim_{A\to\infty} \left( \ln \right(\frac A {A+3}\left) - \ln \right(\frac 2 5 \left)\right) $$

$$= \ln 1 - \ln \left(\frac{2}{5}\right)$$

Therefore by Integral test the original series converges.

I was never thought how to find the sum of this type of series. The only way I know how to find the sum of a series is if its geometric. Does anyone know how too find the sum?

  • $\begingroup$ Hint: Use the following identity: $$\frac{3}{n^2+2n}=\frac{3}{2n}-\frac{3}{2(n+2)}.$$ $\endgroup$
    – M. Scarlet
    Mar 27, 2017 at 3:29

4 Answers 4


To compute the sum observe \begin{align} \frac{1}{n(n+2)} = \frac{1}{2n}-\frac{1}{2(n+2)} \end{align} which means \begin{align} \sum^\infty_{n=2}\frac{3}{n^2+2n} =&\ \frac{3}{2}\sum^\infty_{n=2}\left( \frac{1}{n}-\frac{1}{n+2}\right)\\ =&\ \frac{3}{2}\sum^\infty_{n=2}\left(\frac{1}{n}-\frac{1}{n+1}\right)+\frac{3}{2}\sum^\infty_{n=2}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\\ =&\ \frac{1}{2}+\frac{3}{4}. \end{align}

  • $\begingroup$ Is this telescoping? So if i telescope i get this answer? $\endgroup$
    – user349557
    Mar 27, 2017 at 3:32
  • $\begingroup$ Yes, this is telescoping sum. $\endgroup$ Mar 27, 2017 at 3:32

An alternative approach. The given series is absolutely convergent since the main term behaves like $\frac{1}{n^2}$ and $\sum_{n\geq 1}\frac{1}{n^2}$ is convergent by the $p$-test. We have $$ \frac{3}{n(n+3)}= \int_{0}^{1}\left(x^{n-1}-x^{n+2}\right)\,dx \tag{1}$$ hence the whole series equals $$ \int_{0}^{1}\sum_{n\geq 2}\left(x^{n-1}-x^{n+2}\right)\,dx = \int_{0}^{1}(x+x^2+x^3)\,dx = H_4-1=\color{red}{\frac{13}{12}}.\tag{2}$$


An often used technique in this situation is called partial fraction decomposition; set:

$$\frac{3}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2}$$

and solve for $A$ and $B$. You will find $A = 3/2$ and $B = -3/2$, which means:

$$\frac{3}{n(n+2)} = \frac{3}{2n} - \frac{3}{2(n+2)}$$

The series now "telescopes" as others have just posted. If you have never seen this method before, here is how you actually solve for $A$ and $B$:

Multiplying the first equation by $n(n+2)$ on both sides yields:

$$3 = A(n+2) + Bn \implies 0n + 3 = (A+B)n + 2A$$

The $0n$ is included for emphasis, for this means $A = -B$. Now, all there is to do is observe that you have $2A = 3$, i.e., $A = 3/2$, which yields $B = -3/2$ as desired.

Also, for what it's worth: A different way to show that the series converges is to observe that each term is less than $3/n^2$, which means the series is bounded above by $3 \sum_{n>1}1/n^2$, which is just three multiplied by a convergent series (perhaps what you call a $p$-series). Since the series is monotonically increasing and bounded above, it must converge.


Consider the partial sum $$S_p=\sum_{n=2}^p\frac{3}{n^2+3n}=\sum_{n=2}^p\frac{1}{n}-\sum_{n=2}^p\frac{1}{n+3}$$ $$S_p=\left(\frac 12+\frac 13+\frac 14+\frac 15+\cdots+\frac 1p\right)-\left(\frac 15+\frac 16+\frac 17+\frac 18+\cdots+\frac 1{p+3}\right)$$ What is left is $$S_p=\frac{13}{12}-\frac{1}{p+1}-\frac{1}{p+2}-\frac{1}{p+3}=\frac{13}{12}-\frac{3 p^2+12 p+11}{p^3+6 p^2+11 p+6}$$ Now, for large $p$, perform the long division (or Taylor series) to get $$S_p=\frac{13}{12}-\frac{3}{p}+\frac{6}{p^2}-\frac{14}{p^3}+O\left(\frac{1}{p^4}\right)$$ For $p=10$, $S_{10}=\frac{119}{143}\approx 0.8322$ while the expansion gives $\frac{311}{375}\approx 0.8293$.

You also could use the harmonic numbers and arrive to $$S_p=\frac{13}{12}+H_p-H_{p+3}$$ and using the asymptotics $$H_n=\gamma +\log \left({n}\right)+\frac{1}{2 n}-\frac{1}{12 n^2}+O\left(\frac{1}{p^4}\right)$$ and arrive to the same result.


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