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Solution to puzzle 60: Sum of reciprocals

Find the limit as n tends to infinity of 1/(n+1) + 1/(n+2) + ... + 1/(2n).


By Riemann sum

We have 1/(n+1) + 1/(n+2) + ... + 1/(2n) = Sum for i = 1 to n (1/(1 + i/n) * 1/n).

This corresponds to a Riemann sum for the function f(x) = 1/(1 + x), over the closed interval [0, 1], using n subintervals of equal width.  The graph below illustrates the lower Riemann sum on this interval for n = 8.

Graph illustrating lower Riemann sum for f(x) = 1/(1+x), over [0,1], with n = 8.

Therefore the limit as n tends to infinity of 1/(n+1) + 1/(n+2) + ... + 1/(2n) = integral from 0 to 1 of 1/(1+x) = ln 2.
 

By Maclaurin series

Consider

1/(n+1) + 1/(n+2) + ... + 1/(2n) = ... = 1 - 1/2 + ... + 1/(2n-1) - 1/(2n).

The Maclaurin series for ln(1 + x), valid for −1 < x less than or equal to 1, is:

ln(1 + x) = x - x^2/2 + x^3/3 - ...

The result now follows by setting x = 1.


Remark

More generally, if p and q are positive integers with p < q, then

More generally, the limit as n tends to infinity of 1/(pn+1) + 1/(pn+2) + ... + 1/(qn) = ln(q/p)


Further reading

  1. Riemann Sum Applet

Source: Traditional

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