[RESOLVED] Limit of sum of series, proof required

By numerical analysis, I have determined that the limit as n tends to infinity of the sum over i = 0 to n-1 of 1/(n+i) is 0.693147 (i.e. ln(2))

Please can someone provide a step by step proof that the limit is in fact ln(2) (natural log of 2).

Thanks

Rob

I think you mean the alternating harmonic series converges to ln(2)--that is, the sum of k=1 to infinity of (-1)^k+1/k. As n goes to infinity, 1/(n+i) goes to zero so the sequence you've stated converges to zero since each term goes to zero.

To prove that, find the Taylor Series of ln(1+x) about 0 [which is remarkably simple], and plug in x=1 to find a series for ln(2). The two should be identical.

The sum I am interested in has n terms in it, the terms are 1/n, 1/(n+1), 1/(n+2)...1/(n+n-1), and I'm interested in the limit of this sum as n becomes large (tends to infinity). Now each term is larger than 1/2n and there are n terms, so a lower bound on the sum is 0.5. Computing the sum for large n gives a value of 0.693. I surmise that as n tends to infinity, the sum tends to ln(2) but need a proof of that.

Thanks

Rob

Thanks for the tip on looking at expansions of ln(x).

Rob

The nth term of the sequence is:

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

and you want to prove that

L = lim(n=>infinity){s}_n = ln(2)

Write it as follows:

s_n = [1 + 1/2 + 1/3 + ... + 1/n + 1/(n+1) + 1/(n+2) + ... 1/(2n+1) - ln(2n+1)] - [1 + 1/2 + 1/3 + ... + 1/n + 1/(n+1) + 1/(n+2) + ... 1/(n-1) - ln(n-1)] + ln(2n-1) - ln(n-1)

It just turns out that the 2 sums enclosed in square brackets constitute a sequence of known limit:

lim(n=>infinity) [1 + 1/2 + 1/3 + ... + 1/n - ln(n)] = 0.5772... = g ("gamma")

and is called the "constant of Euler-Mascheroni", see http://mathworld.wolfram.com/Euler-M...iConstant.html

Therefore, using the fact that the limit of a sum is the sum of limits, the properties of logarithms and the fact that the limit of a logarithm is the logarithm of the limit,

lim(n=>infinity)s_n = g - g - lim[ln(2n-1)] + lim[ln(n-1)] = lim{ln[(n-1)/(2n-1)]} = ln{lim[(n-1)/(2n-1)]} = ln(1/2)

Many thanks.

In the end I got there a slightly different way myself.

Let s(k) be the sum of the k terms 1/k, 1/k+1, 1/k+2..1/(2k-1)
So s(1) = 1, s(2) = 1/2 + 1/3, s(3) = 1/3 +1/4 +1/5 etc.

s(k+1) = s(k) -1/k +1/2k + 1/(2k+1)
= s(k) -1/2k + 1/(2k+1)

The value I'm interested in is s(infinity) which can be obtained from s(1) by applying the increment above for values 1..inifinity hence:

s(infinity) = s(1) + sum(x=1..infinity) of (-1/2x + 1/(2x+1))
= sum(x=1..infinity) of (((-1)^(x+1))/x )

which is the alternating harmonic series = ln(2)

:)

Rob