It can be proven by induction that n∑k=11k2≤2−1n
From here, we can easily acquire the upper bound of the sum 99∑k=11k2
letting n=100.
However, I am not quite sure about the lower bound. The standard method of constructing lower rectangles of unit width on the curve y=1x2 yields a lower bound of 0.99, which isn't tight enough unfortunately. What could I do to achieve a sharper bound?
I know that as n→∞, the sum converges to π26, but I feel that proving that result, just for this inequality, is a bit much.
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