JEEOlympiad
The infinite sum sum_{n=1}^{infinity} n/(n^4 + n^2 + 1) equals m/n in…
Question
The infinite sum sum_{n=1}^{infinity} n/(n^4 + n^2 + 1) equals m/n in lowest terms. Find m + n.
✓ Verified answer: 3checked by our engine — not a guess
Step-by-step solution
Factor n^4 + n^2 + 1 = (n^2 - n + 1)(n^2 + n + 1).
Note (n^2 + n + 1) - (n^2 - n + 1) = 2n, so
n/(n^4+n^2+1) = (1/2)[1/(n^2 - n + 1) - 1/(n^2 + n + 1)].
Since n^2 + n + 1 = (n+1)^2 - (n+1) + 1, the bracket telescopes.
Partial sum to N = (1/2)[1/(1^2-1+1) - 1/(N^2+N+1)] = (1/2)[1 - 1/(N^2+N+1)] -> 1/2.
m/n = 1/2, so m + n = 3.
Final answer3
Stuck on a problem like this?
Paste any JEE or NEET question — verified working, a confidence %, and an honest “not sure” instead of a bluff.
Solve my doubt →More Sequences Series solutions
Compute floor(1000 * T), where T = sum_{n=0}^{infinity} (n^2 + 1)/n!…JEE · MathThe sum S = sum_{n=1}^{80} 1/((n+1)*sqrt(n) + n*sqrt(n+1)) equals m/n…JEE · MathLet F_1 = 1, F_2 = 1 and F_{n+2} = F_{n+1} + F_n be the Fibonacci…JEE · MathThe infinite sum sum_{n=1}^{infinity} 1/(n(n+2)(n+4)) equals m/n in…JEE · MathCompute sum_{n=1}^{1023} floor(log_2 n), where floor is the greatest…JEE · Math