Question #e5cde

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Answer 1 · 2016-03-30T13:49:02

Your sum can be expressed as a Riemann Sum. Hence

$\frac 1{n+1}+...+\frac 1{2n}=\int_n^{2n}\frac {1}{x}dx=ln(2n)-ln(n)=ln(2)~~0.693$

Also notice that

$1/2=\frac n{2n}\le\frac1{n+1}+\frac1{n+2}+\ldots+\frac1{2n}\le\frac n{n+1}$

As well as

$\sum_{k=1}^n\frac1{n+k}\leq 1 - \sum_{k=1}^n\frac{k}{2n^2}=1-\frac{n(n+1)}{4n^2}=3/4-\frac1{4n}$

because

$\frac1{n+k}\leq\frac1n-\frac{k}{2n^2}$

Answer 2 · 2016-03-30T13:50:08

The limit is $ln2$ , but I don't see how to get it using the sandwich theorem.

We want $lim_(nrarroo)sum_(i=1)^n1/(n+i)$

$lim_(nrarroo)sum_(i=1)^n1/(1+(i/n))1/n$

This is an integral.

$int_a^b f(x) dx = lim_(nrarroo)sum_(i=1)^n f(x_i) Delta x$

Where $Delta x = (b-a)/n$ $" "$ and $" "$ $x_i = a+iDeltax$

Here we can have $Delta x = 1/n$
and use either

$a=1$ so $b= 2$ and $x_i = 1+i/n$
In which case $f(x) = 1/x$ and we have $int_1^2 1/x dx = ln2$

Or we can use

$a=0$ so $b=1$ and $x_i = i/n$
In which case $f(x) = 1/(1+x)$ and we have $int_0^1 1/(1+x) dx = ln2$