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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#