Question

The sum of the series \(\sum\limits_{n=1}^{\infin}\frac{2n+1}{(n^2+1)(n^2+2n+2)}\) is equal to _________.(rounded off to two decimal places)

Answer 1

Correct Answer: 0.49 - 0.51

To solve the infinite series \(\sum\limits_{n=1}^{\infin}\frac{2n+1}{(n^2+1)(n^2+2n+2)}\), we begin by simplifying the expression. Notice that the denominator can be factored into two quadratic terms:

\(n^2 + 2n + 2 = (n+1)^2 + 1\)

This hints at a telescoping approach. Let's attempt partial fraction decomposition:

\(\frac{2n+1}{(n^2+1)(n^2+2n+2)} = \frac{A}{n^2+1} + \frac{B}{n^2+2n+2}\)

Multiplying through by the common denominator gives:

\(2n+1 = A(n^2+2n+2) + B(n^2+1)\)

Expanding, we have:

\(2n+1 = (A+B)n^2 + (2A)n + (2A + B)\)

Matching coefficients, we find:

• Nodal term: \(A+B=0\)
• Linear term: \(2A=2\) → \(A=1\)
• Constant term: \(2A+B=1\)

Solving these, \(A=1\) and \(B=-1\).

Substituting back, the series becomes:

\(\sum\limits_{n=1}^{\infin}\left(\frac{1}{n^2+1} - \frac{1}{n^2+2n+2}\right)\)

Now, recognize the telescoping nature of the series:

\(\left(\frac{1}{1^2+1} - \frac{1}{2^2+2}\right) + \left(\frac{1}{2^2+1} - \frac{1}{3^2+3}\right) + \ldots\)

Most intermediate terms cancel out, leaving:

\(\frac{1}{1^2+1} = \frac{1}{2}\)

To confirm, we note this fits within the given range [0.49, 0.51]. Thus, the sum of the series is 0.50.