Numerical: infinite series MCQ — 2 Practice Questions with Answers

Explanation

The denominator of the nth term is the sum of the first n natural numbers, n(n+1)/2. So each term is 2/[n(n+1)], which can be written as 2(1/n - 1/(n+1)). The infinite series therefore telescopes: 2[(1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ...]. All middle terms cancel, leaving 2. The values 1/2, 1, and 3/2 are partial or underestimated values, not the infinite sum.

Explanation

The general term is \frac{2n}{(2n+1)!}. Since 2n=(2n+1)-1, it becomes \frac{1}{(2n)!}-\frac{1}{(2n+1)!}. Summing from n=1 gives \left(\frac{1}{2!}+\frac{1}{4!}+\cdots\right)-\left(\frac{1}{3!}+\frac{1}{5!}+\cdots\right). This equals (\cosh 1-1)-(\sinh 1-1)=\cosh 1-\sinh 1=e^{-1}. Hence e, 2e, and e^2 are too large and do not match the alternating even-odd factorial difference.