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Show that for all large positive integers n the sum 1/(n+1) + 1/(n+2) + 1/(n+3) +...

Show that for all large positive integers n the sum 1/(n+1) + 1/(n+2) + 1/(n+3) + ... + 1/(2n) is approximately equal to 0.693.

I am trying to solve this problem by setting the sigma summation from k = n + k to 2n of 1/j to try to make a harmonic sum but is not working. I let j be n + k so it matches the harmonic sum definition of 1/k

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Answer #1

`Hey,

Note: Brother in case of any queries, just comment in box I would be very happy to assist all your queries

We can rewrite the equation as

Also we know that

Sum of harmonic series with n terms is

Hn=ln(n)+0.5772156649-----eq1

So,

H2n=ln(2n)+0.5772156649-------eq2

Subtract eq1 and eq2

H2n-Hn=ln(2)~0.6931

Kindly revert for any queries

Thanks.

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