Vtyjkyuk

$$lim_ntoinftysumlimits_0leqslantkleqslant2n frackk+n^2$$

I can't figure out the "right" function for this limit. The previous problem was very similar except there was $k^2$ in the denominator, and it wasn't difficult to recognize the Riemann sum for $f(x)=fracxx^2+1$.

Given the limits, this integral sum can be written as $sumf(frac2kn)cdotfrac2n$.

Write the sum as

$$S_nn = frac1nsum_k=0^2n frack/n1 + k/n^2$$

This is not in the form of a Riemann sum. However, there is a trick (requiring justification) where we can reduce the limit to that of a Riemann sum by viewing this in terms of a double sequence

$$S_nm = frac1nsum_k=0^2n frack/n1 + (k/n)(1/m)$$

Clearly, $displaystylelim_m to infty S_nm = frac1nsum_k=0^2n frackn.,$ As the convergence can be shown to be uniform for all $n$, by a well-known theorem for double sequences, we can evaluate as an iterated limit,

$$lim_n to infty S_nn = lim_n to infty lim_m to infty S_nm = lim_n to infty frac1nsum_k=0^2n frackn = int_0^2 x , dx = 2.$$

Justification: Uniform convergence of inner limit

Note that

$$left|frac1nsum_k=0^2n frack/n1 + (k/n)(1/m) - frac1nsum_k=0^2n frackn right|= frac1nsum_k=0^2nfrac(k/n)^2(1/m)1 + (k/n)(1/m) \ leqslant frac1nmsum_k=0^2nleft(fracknright)^2 leqslant frac2n nmleft(frac2nn right)^2 = frac8m$$

Edited after reading the comments.

No need for Riemann sums.

Lower bound:
$$
sum_k=0^2nfrackk+n^2gefrac12,n+n^2sum_k=0^2nk=fracn,(2,n+1)2,n+n^2.
$$
Upper boud:
$$
sum_k=0^2nfrackk+n^2lefrac1n^2sum_k=0^2nk=fracn,(2,n+1)n^2.
$$
However, Riemann sums can be used, but in a more complicated way:
$$
sum_k=0^2nfrackk+n^2=2,n,Bigl(frac12,nsum_k=0^2nfrack/(2,n)k/(2,n)+n/2Bigr)
sim2,nint_0^1fracx,dxx+n/2.
$$