Question

Write \(\sum_{k=1}^{\infty} \frac{6^{k}}{\left(3^{k+1}-2^{k+1}\right)\left(3^{k}-2^{k}\right)}\) as a rational number.

Short answer

\(\frac{1}{2}\)

Step-by-step solution

Identify the pattern

Looking at the given series, there seems to be no evident pattern. But we can see a term, \(3^k\), is common in both the denominator \(3^{k+1}-2^{k+1}\) and \(3^{k}-2^{k}\). We need to rewrite the denominator in such a way that a pattern emerges. The motive should be to convert the series into a geometric series because we know how to sum a geometric series.

Re-arrange equations

After some rearrangement, applying the properties of exponentiation, and simplifying the fractions, we can write \(\frac{6^{k}}{\left(3^{k+1}-2^{k+1}\right)\left(3^{k}-2^{k}\right)}\) as \[\frac{2^{k}*3^{k}}{(3*3^{k}-2*2^{k})(3^{k}-2^{k})}\] Now, we try to separate out \(3^k\) and \(2^k\) from the denominator by dividing both the terms individually. The new equation becomes \[\frac{3^{k}}{3*3^{k}-2*2^{k}}-\frac{2^{k}}{3^{k}-2^{k}}\] Now we have the terms in the form of a telescoping series.

Recognize and evaluate the telescoping series

A telescoping series is a series where most of the terms cancel out and only a finite number of terms are left. The sum of the series can then be found by adding these finite terms. Observe that the term \(\frac{3^{k}}{3*3^{k}-2*2^{k}}\) at \(k=1\) will cancel out with the term \(\frac{2^{k}}{3^{k}-2^{k}}\) at \(k=2\) and so on. Eventually, they will all cancel out save for the first term of \(\frac{3^{k}}{3*3^{k}-2*2^{k}}\) and the last term of \(\frac{2^{k}}{3^{k}-2^{k}}\) as \(k \to \infty\). We can find these two terms to sum them up.

Evaluate the remaining terms and Sum

The first term of the series \(\frac{3^{k}}{3*3^{k}-2*2^{k}}\) where \(k=1\) is \(\frac{1}{2}\) and the last term of the series \(\frac{2^{k}}{3^{k}-2^{k}}\) where \(k \to \infty\) goes to 0 as the value of \(k\) is so large. Thus, the sum of the series is \(\frac{1}{2}-0=\frac{1}{2}\) which is a rational number.