Telescoping sums: fractions, consecutive products, and arithmetic-geometric sums

What makes a sum telescope?

If a term can be written as $a_k=g(k+1)-g(k)$, then

$$\begin{aligned} \sum_{k=1}^n a_k &=\{g(2)-g(1)\}+\{g(3)-g(2)\}+\cdots+\{g(n+1)-g(n)\}\\ &=\boxed{g(n+1)-g(1).} \end{aligned}$$

Every middle term cancels. The function $g$ is a discrete counterpart of an antiderivative: taking its difference recovers the summand.

Telescoping is especially useful when a term has consecutive factors or combines a polynomial with a geometric sequence. It is not the only method for such sums, but it can avoid a long expansion.

Example 1: a fractional sum

Problem. Evaluate

$$\sum_{k=1}^n\frac1{k(k+1)}.$$

Partial fractions give

$$\frac1{k(k+1)}=\frac1k-\frac1{k+1}.$$

Writing a few terms reveals the cancellation:

$$\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1n-\frac1{n+1}\right).$$

Thus

$$\boxed{\sum_{k=1}^n\frac1{k(k+1)}=1-\frac1{n+1}.}$$

In the $g(k+1)-g(k)$ convention, use $g(k)=-1/k$. The negative sign ensures that the difference is taken in the right direction.

Example 2: a product of consecutive integers

Problem. Evaluate

$$\sum_{k=1}^n k(k+1)(k+2)(k+3).$$

Instead of expanding the four factors, compare the product multiplied by the next and previous integers:

$$\begin{aligned} &k(k+1)(k+2)(k+3)(k+4)\\ &\quad-(k-1)k(k+1)(k+2)(k+3)\\ &=k(k+1)(k+2)(k+3)\{(k+4)-(k-1)\}\\ &=5k(k+1)(k+2)(k+3). \end{aligned}$$

Therefore choose

$$g(k)=\frac{(k-1)k(k+1)(k+2)(k+3)}5.$$

Its difference is the summand, so

$$\sum_{k=1}^n k(k+1)(k+2)(k+3)=g(n+1)-g(1).$$

Since $g(1)=0$, the answer is

$$\boxed{\frac{n(n+1)(n+2)(n+3)(n+4)}5.}$$

The general pattern

For $m$ consecutive factors, the same subtraction introduces a factor $m+1$:

$$\boxed{\sum_{k=1}^n k(k+1)\cdots(k+m-1)=\frac{n(n+1)\cdots(n+m)}{m+1}.}$$

Here $m,n$ are positive integers. This is often simpler than using separate formulas for several powers of $k$.

Example 3: an arithmetic sequence times a geometric sequence

Problem. Evaluate

$$\sum_{k=1}^n(2k+1)3^k.$$

A familiar approach is to multiply the whole sum by $3$ and subtract. Telescoping uses the same cancellation locally, by finding $g$ first.

Because the summand is linear in $k$ times $3^k$, try

$$g(k)=(Ak+B)3^k.$$

Then

$$\begin{aligned} g(k+1)-g(k) &=\{A(k+1)+B\}3^{k+1}-(Ak+B)3^k\\ &=(2Ak+3A+2B)3^k. \end{aligned}$$

Equating coefficients with $(2k+1)3^k$ gives $2A=2$, $3A+2B=1$, hence $A=1$, $B=-1$. Thus $g(k)=(k-1)3^k$.

Before summing, check the proposed difference directly:

$$k3^{k+1}-(k-1)3^k=(3k-k+1)3^k=(2k+1)3^k.$$

It is correct, so

$$\boxed{\sum_{k=1}^n(2k+1)3^k=g(n+1)-g(1)=n3^{n+1}.}$$

For $n=1$, both sides give $9$, a quick endpoint check.

How to choose an ansatz in similar problems

For $(pk+q)r^k$ with $r\ne0,1$, trying $g(k)=(Ak+B)r^k$ yields

$$g(k+1)-g(k)=\{(r-1)Ak+rA+(r-1)B\}r^k.$$

Equate the linear and constant coefficients to determine $A,B$. Higher-degree polynomial factors suggest a polynomial of the same degree in the ansatz. If $r=1$, the problem is an ordinary polynomial sum and needs a different form for $g$.

A checking routine that prevents endpoint mistakes

First expand $g(k+1)-g(k)$ to verify the summand. Then write the first two and last two differences, so the surviving endpoints are visible. Finally substitute a small value of $n$ into the answer.

Partial fractions, expansion into power sums, and multiplying the whole sum by the common ratio remain valid alternatives. Choose the route that makes the cancellation easiest to verify.

Related: proofs of the square, cube, and fourth-power sum formulas.