n-1-n-n-2-log-n-1-In-explicit-form-

Question Number 124718 by Dwaipayan Shikari last updated on 05/Dec/20

\underset{n = 1}{\sum^{n}} n^{2} l o g (n + 1) (I n e x p l i c i t f o r m)

Commented by Dwaipayan Shikari last updated on 05/Dec/20

I n {R a m a n u j a n}^{'} s N o t e b o o k

1^{2} l o g (1) + 2^{2} l o g (2) + 3^{2} l o g (3) + 4^{2} l o g (4) + \dots + n^{2} l o g (n)

= \frac{n (n + 1) (2 n + 1)}{6} l o g (n) - \frac{n^{3}}{9} + \frac{ζ (3)}{4 π^{2}} + \frac{n}{12} - \frac{1}{360 n} + \dots + C

B u t i {h a v e n}^{'} t f o u n d a n y p r o o f