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Compute $$\lim_{n \to \infty} \sum_{k=1}^{n} \frac{n^2 + k}{n^3 + k}$$

I tried: $$ 1 \xleftarrow{n \to \infty} n \cdot \frac{n^2+1}{n^3 + n} \le \sum_{k=1}^{n} \frac{n^2 + k}{n^3 + k} \le n \cdot \frac{n^2 + n}{n^3 +1} \xrightarrow{n \to \infty} 1$$

Hence:$$\lim_{n \to \infty} \sum_{k=1}^{n} \frac{n^2 + k}{n^3 + k} = 1$$

I have no answer to that task and wolfram alpha didn't help me. I will grateful if you could check it.

Thomas
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2 Answers2

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Your answer is correct. That's a nice application of the sandwich theorem (plus some artistic equation writing in the second line).

Lord Soth
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  • Paraphrasing a line from a Blackadder episode, I'm making a note of that artistic equation writing. I want to use it in conversation. – Gunnar Þór Magnússon Aug 15 '13 at 19:40
  • @GunnarMagnusson Non-nativity often results in the "invention" of new phrases. "Artistic equation writing" and "non-nativity" may be examples of this phenomenon. – Lord Soth Aug 15 '13 at 19:52
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Your answer is very nice. Here is an alternate method in support: $$ \lim_{n\to\infty}\sum_{k=1}^n\frac{n^2+k}{n^3+k} =\lim_{n\to\infty}\sum_{k=1}^n\frac{1+k/n^2}{1+k/n^3}\frac1n $$ For $0\le k\le n$, $$ 1\le\frac{1+k/n^2}{1+k/n^3}\le1+\frac1n $$ Thus, the right hand side is between $1$ and $1+\frac1n$ times a Riemann Sum $\left(x=\frac kn,\,\mathrm{d}x=\frac1n\right)$ for $$ \int_0^11\,\mathrm{d}x=1 $$ As $n\to\infty$, the right hand side is arbitrarily close to $1$.

robjohn
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    Is it really a Riemann sum? – Pedro Aug 15 '13 at 19:47
  • @PeterTamaroff: I guess it depends on your formulation of Riemann Sums. In any case, I have altered my approach to fit almost any formulation, but at the cost of perhaps getting closer to the OP's answer (i.e. applying the Squeeze Theorem). – robjohn Aug 15 '13 at 20:06