Telescoping sums and products

Algebra

Expressing sums or products as ‘telescoping’ ones is a very standard trick which I have seen to be quite enjoyed by high school students, although it is hardly introduced at the ordinary high school level (at least in India). Despite its simplicity it is widely applicable and used regularly in the academia.

Let me illustrate the main idea via the following simple examples:

Question. What is the value of the sum \displaystyle\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \cdots + \frac{1}{9900} ?

Solution. You can easily recognise that the n-th summand above is nothing but \displaystyle\frac{1}{n(n+1)} which we can also write as

\displaystyle\frac{1}{n(n+1)} = \frac{(n+1) - n}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1}.

Now the trick is to observe that after writing the summands in the above manner, successive terms get cancelled out:

\displaystyle\frac{1}{1\times 2} + \frac{1}{2\times 3} + \frac{1}{3\times 4} + \cdots + \frac{1}{99\times 100} = 1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} +\frac{1}{3} - \cdots + \frac{1}{99} - \frac{1}{100}.

We can see that only the first and the last terms survive and hence the desired sum equals \displaystyle 1- \frac{1}{100} =\frac{99}{100}. We often say that the above sum telescopes to \displaystyle 1 - \frac{1}{100}.

The same trick also applies to products, e.g., consider the following question.

Question. Evaluate \displaystyle \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{3}\right)\cdots \left(1 - \frac{1}{n}\right).

Solution. This is even simpler than the previous question — just note how consecutive denominator and numerator cancel each other:

\displaystyle \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{3}\right)\cdots \left(1 - \frac{1}{n}\right) = \frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \cdots \frac{n-1}{n} = \frac{1}{n}.

If you have understood the main idea, try the following problems which are more challenging (and perhaps more interesting).

#1. Prove the identity \displaystyle\prod_{n=1}^\infty \cos\left( \frac{x}{2^n}\right) = \frac{\sin x}{x}.

#2. Simplify: \displaystyle \cos\left( \frac{2\pi}{2^n-1}\right)\cos\left( \frac{4\pi}{2^n-1}\right) \cdots \cos\left( \frac{2^n\pi}{2^n-1}\right).

#3. Show that \displaystyle \frac{1}{\cos 0^\circ \cos 1^\circ} + \frac{1}{\cos 1^\circ \cos 2^\circ} + \cdots + \frac{1}{\cos 88^\circ \cos 89^\circ} = \frac{\cos 1^\circ}{\sin^2 1^\circ}.

#4. Prove that \displaystyle \frac{\sin x}{\cos x} + \frac{\sin (2x)}{\cos^2 x} + \cdots+ \frac{\sin (nx)}{\cos^n x} = \cot x - \frac{\cos ((n+1)x)}{\sin x \cos^n x}.

#5. Evaluate the series: \displaystyle \sum_{k=0}^\infty \cot^{-1} (k^2+k+1).

#6. Evaluate the series: \displaystyle \sum_{k=1}^\infty \tan^{-1} \frac{1}{2k^2}.

#7. Find a formula for \displaystyle \sum_{k=1}^n \cos (kx). (Treat the case x=2m\pi, m\in \mathbb{Z} separately.)

#8. Simplify: \displaystyle \frac{1}{\cos a - \cos 3a} + \frac{1}{\cos a - \cos 5a} + \cdots + \frac{1}{\cos a - \cos (2n+1)a}.

#9. Evaluate the series: \displaystyle \sum_{n=1}^\infty \frac{1}{2^n}\tan \frac{a}{2^n}, assuming a \neq k\pi for any k \in \mathbb{Z}.

#10. Show that \displaystyle \sum_{n=1}^\infty 3^{n-1}\sin^3 \frac{a}{3^n} = \frac{a-\sin a}{4}.

#11. Simplify: \displaystyle \sum_{k=1}^n \frac{1}{\sin (2^k x)}, assuming that x is not of the form k\pi/2^m for any m \in\mathbb{Z}.

#12. Prove that the average of the numbers n\sin n^\circ where (n = 2,4, 6\dots, 180) is \cot 1^\circ .

#13. Evaluate: \displaystyle (1 - \cot 1^\circ)(1 - \cot 2^\circ) \cdots (1 - \cot 44^\circ).

#14. Simplify: \displaystyle \frac{\tan 1}{\cos 2} + \frac{\tan 2}{\cos 4} + \frac{\tan 4}{\cos 8}+\cdots + \frac{\tan 2^n}{\cos 2^{n+1}}. (Here the angles are in radian.)

#15. Evaluate: \displaystyle \prod_{k=1}^n \left(1 - \tan^2 \left(\frac{2^k \pi}{2^n+1} \right)\right).

#16. Show that \displaystyle \left( \frac{1}{2} + \cos\frac{\pi}{20}\right)\left( \frac{1}{2} + \cos\frac{3\pi}{20}\right)\left( \frac{1}{2} + \cos\frac{9\pi}{20}\right)\left( \frac{1}{2} + \cos\frac{27\pi}{20}\right) = \frac{1}{16}. Generalize.

#17. If x is not of the form 2^{k+1} (\pi/3 + \ell \pi) for k=1,2,\dots,n and for any \ell\in\mathbb{Z}, show that

\displaystyle\prod_{k=1}^n \left(1-2\cos\frac{x}{2^k}\right) =(-1)^n \frac{1+2\cos x}{1+2\cos (x/2^n)}.

#18. Prove that \displaystyle \prod_{k=1}^n \left(1 + 2\cos \left(\frac{3^k \cdot 2\pi}{3^n+1} \right)\right) = 1.

The above problems in a pdf: click here.

For solutions to the above problems, click here. (It is from an old diary of mine, I apologize for the untidiness.)

If you have more such problems, I encourage you to share them by commenting below.

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