I am familiar with the definition of Riemann sums and how they are used to evaluate definite integrals. But I am completely stuck with this expression. I can't seem to extract any patterns that relate to Riemann sums (other than the limit and summation term). Am I supposed to assume an interval in terms of $k$ and try to formulate an expression of the form $f(x_i^*)\Delta x$? Any hints would be greatly appreciated!
Solve $\lim_{n \to \infty} \sum_{k=1}^{n}\frac{k^2}{2n^3 + k^3}$ using Riemann sums.
calculusreal-analysisriemann sumriemann-integration
Related Solutions
Note that the mesh spacing is not uniform. The length of the $i$'th subinterval is $x_i-x_{i-1}=2^{i/n}-2^{(i-1)/n}$.
Then, we can write
$$\begin{align} L(f,P_n)&=\sum_{i=1}^n \frac{1}{x_{i-1}}(x_i-x_{i-1})\\\\ &=\sum_{i=1}^n \left(\frac{1}{2^{(i-1)/n}}\right)\left(2^{i/n}-2^{(i-1)/n}\right)\\\\ &=\sum_{i=1}^n\left(2^{1/n}-1\right)\\\\ &=n\left(2^{1/n}-1\right)\\\\ &=n\left(e^{\frac1n\log(2)}-1\right)\\\\ &=\log(2)+O\left(\frac1n\right)\tag 1 \end{align}$$
and
$$\begin{align} U(f,P_n)&=\sum_{i=1}^n \frac{1}{x_{i}}(x_i-x_{i-1})\\\\ &=\sum_{i=1}^n \left(\frac{1}{2^{i/n}}\right)\left(2^{i/n}-2^{(i-1)/n}\right)\\\\ &=\sum_{i=1}^n\left(2^{1/n}-1\right)\\\\ &=n\left(1-2^{-1/n}\right)\\\\ &=n\left(1-e^{-\frac1n\log(2)}\right)\\\\ &=\log(2)+O\left(\frac1n\right)\tag 2 \end{align}$$
Taking the limit as $n\to \infty$ of $(1)$ and $(2)$, we find
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}L(f,P_n)=\lim_{n\to \infty}U(f,P_n)=\log(2)}$$
as was to be shown!
If $f \colon [a,b] \rightarrow \mathbb{R}$ is continuous, then $f$ is Riemann-integrable over $[a,b]$ and then it is enough to take one shrinking partition, i.e. $(P_n)_{n \in \mathbb{N}}$ with $m(P_n) \rightarrow 0$ to evaluate the limes $\lim_{n \rightarrow \infty} R(f,P_n)$.
Proof: $f$ is uniformly contiuous on $[a,b]$, i.e. for any $\varepsilon>0$ there exists $\delta >0$ with $|x-y|<\delta$, $x,y \in [a,b]$ implies $|f(x)-f(y)| < \delta$. Now let $P =(t_0,\ldots,t_m)$ be any Partition with $\mu(P)<\delta$ and take $n \in \mathbb{N}$ with $m(P_n)< \delta$, write $P_n=(s_0,\ldots,s_l)$, and $|R(f,P_n) - I| < \varepsilon$. We have $$R(f,P_n) = \sum_{k=0}^{l+1} f(\widetilde{s}_k) (s_{k+1}-s_{k})$$ with $\widetilde{s}_k \in [s_k,s_{k+1}]$. We can refine the sum, such that $m=l$ and $s_i =t_i$, and similiar for $R(f,P)$. We may loose the property that $\widetilde{s}_k \in [s_k,s_{k+1}]$. However, if $s_i \le t_k < s_{i+1}$, then $|\widetilde{s_i}-\widetilde{t}_k| < \delta$. Thus the new refinement has the property that the chosen intermediate points for both partitions have distance less then $\delta$. Then we have \begin{align} \tag{1}|R(f,P_n) - R(f,P)| \le \sum_{k=0}^{k-1} |f(\widetilde{s_k}) - f(\widetilde{t_k})| (s_{k+1}-s_k) &< \varepsilon \sum_{k=0}^{k-1} (s_{k+1}-s_k) \\ & \le \varepsilon (b-a). \end{align} Thus, we find $|R(f,P) -I| < \varepsilon(1+b-a)$. The line (1) shows also (by modifing the notation), that $(R(f,P))_{P \text{ Partition}}$ is a Cauchy-net, i.e. convergent, because $\mathbb{R}$ is complete.
The integral definition, where we may only took one sequence of shrinking partition, is called Darboux integral. See also here, where you can find also the relation to Riemann integration.
Best Answer
Hint
$$\sum_{k=1}^n\frac{k^2}{2n^3+k^3}=\frac{1}{n}\sum_{k=1}^n\frac{(k/n)^2}{2+(k/n)^3}.$$