Dgdrxrt

$$lim _nto infty sum _k=1^nfrac1n+k+frackn^2$$ I unsuccessfully tried to find two different Riemann Sums converging to the same value close to the given sum so I could use the Squeeze Theorem. Is there any other way to solve this?

It is easy to see that

$$ beginalign
left |sum_k=1^n frac1n+k+k/n^2-sum_k=1^n frac1n+kright|&=frac1n^2sum_k=1^n frack(n+k+k/n^2)(n+k)\\
&le frac1n^2sum_k=1^n frac1ktag 1
endalign$$

Then, using $sum_k=1^nfrac1k =gamma+log(n) +Oleft(frac1nright)$, we see that the limit of the left-hand side of $(1)$ as $nto infty$ is $0$.

Can you finish now?

Writing this as

$$lim_n to inftyS_nn = lim_n to inftyfrac1nsum _k=1^nfrac11+frackn+fracknfrac1n^2
$$

a technique that often works is to evaluate the double limit

$$lim_m to infty lim_n to infty S_mn = lim_m to infty lim_n to infty frac1nsum _k=1^nfrac11+frackn+fracknfrac1m^2 = lim_m to inftyint_0^1 fracdx1 +x + x/m^2 = int_0^1 fracdx1 +x
$$

where the last step is justified by DCT.

We can justify $ lim_n to infty S_nn = lim_m to infty lim_n to infty S_mn = log 2$ by showing one of the iterated limits exhibits uniform convergence.

An example with more details is given here. The argument for justifying the use of the double limit in this case will be similar.

$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
&bbox[10px,#ffd]lim _n to infty sum _k = 1^n1 over n + k + k/n^2 =
lim _n to infty bracksn^2 over n^2 + 1
sum _k = 1^n1 over k + n^3/parsn^2 + 1
\[5mm] = &
lim _n to infty bracesn^2 over n^2 + 1
sum _k = 1^inftybracks1 over k + n^3/parsn^2 + 1 -
1 over k + n^3/parsn^2 + 1 + n
\[5mm] = &
lim _n to infty bracesn^2 over n^2 + 1
bracksH_large n^3/parsn^2 + 1 + n -
H_large n^3/parsn^2 + 1
endalign
where $dsH_z$ is a Harmonic Number.

In using the
$dsH_z$ asymptotic behavior as $dsvertsz to infty$, it's straightforward found:

$$
bbxlim _n to infty sum _k = 1^n1 over n + k + k/n^2 = lnpars2
$$

I thought it might be instructive to present a method that relies only on the squeeze theorem. To that end, we proceed.

First, it is trivial to see that

$$sum_k=1^n frac1n+k+k/n^2le sum_k=1^n frac1n+ktag 1$$

Second, note that we have

$$beginalign
sum_k=1^n frac1n+k+k/n^2&=sum_k=1^n frac1(n+k)left(1+frack/n^2n+kright)\\
&ge sum_k=1^n frac1(n+k)left(1-frack/n^2n+kright)\\
&ge left(1-frac1n^2right)sum_k=1^n frac1n+ktag2
endalign$$

Putting $(1)$ and $(2)$ together reveals

$$left(1-frac1n^2right)sum_k=1^n frac1n+kle sum_k=1^n frac1n+k+k/n^2le sum_k=1^n frac1n+k$$

whence application of the squeeze theorem yields the coveted limit

$$lim_nto inftysum_k=1^n frac1n+k+k/n^2=log(2)$$

as expected!

The sum under limit is not a Riemann sum, but it differs from a Riemann sum by negligible amount.

To setup things let's note that by definition of Riemann integral we have $$int_0^1f(x),dx=lim_ntoinfty frac1nsum_k=1^nf(t_k),,frack-1nleq t_kleq frack n $$ It is easy to guess that the function $f$ involved here is given by $f(x) =1/(1+x)$ and let's choose $$t_1=0,t_k=frack-1n+frack-1n^3,k=2,3dots,n$$ The corresponding Riemann sum is $$S_n=frac1n+frac1nsum_k=2^ndfrac11+dfrack-1n+dfrack-1n^3$$ which can be rewritten as $$S_n=frac1n+sum_k=1^n-1frac1n+k+(k/n^2)$$ If the sum in question is denoted by $S'_n$ then we can see that $$S_n-S'_n=frac1n-frac1n+n+(1/n^2)$$ and clearly the above expression tends to $0$ as $ntoinfty $. The Riemann sum $S_n$ tends to $int_0^1dx/(1+x)=log 2$ and hence the given sum $S'_n$ also tends to $log 2$.

The same technique has been used in this answer in a simpler manner.