The Sum of the Series 1/(1·2) − 1/(2·3) + 1/(3·4) − ⋯ up to Infinity Is

A standard sequences and series problem, not tied to a specific exam paper.

Problem

The sum of the series $\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}-\cdots$ up to infinity is:

Key insight. $\dfrac{1}{n(n+1)}=\dfrac1n-\dfrac{1}{n+1}$ splits every term of this alternating series into two ordinary reciprocals. Grouping all the “$\frac1n$” pieces together and all the “$\frac{1}{n+1}$” pieces together turns the whole thing into two separate alternating-harmonic-type sums, each of which is a shifted copy of the well-known expansion $\log_e(1+x)$ at $x=1$.

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Approach

Split the general term $\dfrac{(-1)^{n+1}}{n(n+1)}$ using partial fractions into $\dfrac1n-\dfrac1{n+1}$. Separate the resulting sum into the alternating series $\sum\frac{(-1)^{n+1}}{n}$ (which equals $\log_e2$, from the standard expansion) and a second alternating series that’s a shifted version of the same sum. Combine both pieces to reach a closed form.

Solution

Step 1 — Apply partial fractions to the general term

$$\frac{1}{n(n+1)} = \frac1n-\frac{1}{n+1}$$

So the series becomes:

$$S = \sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac1n-\frac1{n+1}\right)$$

Step 2 — Separate into two sums

$$S = \underbrace{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}}_{=\;S_1} \;-\; \underbrace{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n+1}}_{=\;S_2}$$

Step 3 — Recognise S₁ from the standard log(1+x) expansion

$$\log_e(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$

Setting $x=1$:

$$S_1 = 1-\frac12+\frac13-\frac14+\cdots = \log_e2$$

Step 4 — Evaluate S₂ by re-indexing

$$S_2 = \frac12-\frac13+\frac14-\cdots$$

This is $S_1$ with its leading term ($1$) removed and every sign flipped once more — precisely, $S_2 = 1-S_1 = 1-\log_e2$.

Step 5 — Combine

$$S = S_1-S_2 = \log_e2-(1-\log_e2) = 2\log_e2-1$$

Step 6 — Match against the given options

$$2\log_e2-1 = \log_e4-\log_ee = \log_e\left(\frac4e\right)$$

This matches option (B).

Answer

Common mistakes

• Writing out only a few terms and trying to spot a numeric pattern. The series doesn’t telescope in the usual sense (adjacent terms don’t cancel directly) — recognising the connection to the $\log_e(1+x)$ expansion is what actually resolves it, not pattern-spotting from partial sums.
• Getting the sign wrong when re-indexing $S_2$. $S_2=\frac12-\frac13+\frac14-\cdots$ starts with a positive $\frac12$, so writing it as $-(1-\log_e2)$ instead of $+(1-\log_e2)$ flips the final answer’s sign entirely.

Practise next

• Find the sum of $\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}-\cdots$ to infinity, using the same partial-fraction splitting approach.