Find the series: \sum_{n=1}^\infty((1+1/n)^n-e)

Jamarion Kerr 2022-01-22 Answered
Find the series:
n=1((1+1n)ne)
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Expert Answer

immablondevl
Answered 2022-01-23 Author has 11 answers
Take the negative of that series to make the terms positive and limit compare it with 1n
limne(1+1n)n1n00
limn(1+1n)n(log(1+1n)1n+1)1n2
which we get from LHopital
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gekraamdbk
Answered 2022-01-24 Author has 13 answers
We first observe, that
(1+1n)ne=enln(1+1n)e
Elementary calculus arguments can show that the function xln(1+1x) is monotone increasing on (0,) and that its range has 1 as a supremum and has 0 as an infimum.
We now invoke the Mean Value Theorem to conclude that for every nN there is a real ζn between 1 and nln(1+1n) such that
e(1+1n)n=eζn(1nln(1+1n))
From what we know about xln(1+1x), we can conclude that
e0(1nln(1+1n))e(1+1n)ne1(1nln(1+1n))
Now we use the inequalities
12(1+x)1xln(1+1x)1x+1
which is true for all positive x to conclude
12(1+n)e(1+1n)ne1+n
which will carry the rest of the argument.
The real crux to this argument are the inequalities
12(1+x)1xln(1+1x)1x+1
The right-hand inequality is equivalent to the more familiar inequality
ln(1+x)x
which I can derive if needed.
The left-hand inequality is equivalent to the slightly stronger inequality
ln(1+x)xx22(1+x)
which is however established in very much the same way as the weaker inequality.
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