Find the sum of the series: sum from n=1 to infinity
Find the sum of the series: $$ \sum_{n=1}^{\infty} \frac{(-1)^n(2n+1)+1}{n(n+1)}. $$
Explanation of Method
Splitting the numerator into $(-1)^n(2n+1)$ plus $1$ separates the series into a telescoping piece $\sum1/n(n+1)$ and an alternating piece $\sum(-1)^n(2n+1)/n(n+1)$. Both pieces have clean partial-fraction decompositions, and the alternating piece reduces directly to (shifted copies of) the alternating harmonic series, whose sum is the classical constant $\ln2$.
- Split the term into a telescoping part and an alternating part.
- Sum the telescoping part $\sum1/n(n+1)$ via partial fractions.
- Sum the alternating part via partial fractions and the alternating harmonic series.
Step 1 · Telescoping part
$\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}$, so $\displaystyle\sum_{n=1}^{\infty}\frac1{n(n+1)}=\lim_{N\to\infty}\left(1-\frac1{N+1}\right)=1$.
Step 2 · Alternating part
$\dfrac{2n+1}{n(n+1)}=\dfrac1n+\dfrac1{n+1}$, so $$ \sum_{n=1}^{\infty}\frac{(-1)^n(2n+1)}{n(n+1)} = \sum_{n=1}^{\infty}\frac{(-1)^n}n + \sum_{n=1}^{\infty}\frac{(-1)^n}{n+1} = -\ln2 + (\ln2-1) = -1, $$ using $\sum(-1)^n/n=-\ln2$ and reindexing $\sum(-1)^n/(n+1)=\sum_{k\ge2}(-1)^{k-1}/k=\ln2-1$.
Step 3 · Combine
$$ \sum_{n=1}^{\infty}\frac{(-1)^n(2n+1)+1}{n(n+1)} = -1 + 1 = 0. $$ Verified numerically: partial sums $S_1=-1$, $S_2=0$, $S_3=-\frac12$, $S_4=0$, oscillating toward $0$.
Summary Table
| Component | Value |
|---|---|
| $\sum1/n(n+1)$ | $1$ |
| $\sum(-1)^n(2n+1)/n(n+1)$ | $-1$ |
| Total | $0$ |
Additional Notes
- This problem cleanly combines two techniques seen throughout the series (telescoping, from Problems 149 and earlier, and the alternating harmonic series' sum $\ln2$, first used explicitly in Problem 128's boundary case) into a single computation.
- The final cancellation to exactly $0$ is not a coincidence of rounding but an exact algebraic identity, since both partial-fraction decompositions produce rational telescoping structures alongside the transcendental constant $\ln2$, which cancels perfectly between the two reindexed alternating sums.
- This is a useful template for a broader class of problems: any rational function of $n$ multiplied by $(-1)^n$ can be decomposed via partial fractions into shifted copies of the alternating harmonic series, always summable in closed form using $\ln2$.
Result
The series sums to: