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On my sample calculus mid-term exam, there is a weird question that asks us to calculate the limit that has an infinite term:

$\displaystyle\lim_{n \to \infty} {\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}}$

I am not sure what kind of technique should be used here.

Giving that this is mid-term of the first-year university calculus exam, we have only learned L'hopital's Rule and other basic techniques. But I don't see any technique that can fit to solve this problem

Thanks!

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3 Answers 3

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Note that $$\displaystyle\lim_{n \to \infty} {\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}}=\displaystyle\lim_{n \to \infty} \frac{1}{n}{\sum_{k=1}^n\frac{1}{1+\frac{k}{n}}}$$

This is a Riemann Sum for $f(x)=\frac{1}{1+x}$ over the interval $[0,1]$, with the satndard partition $x_i=\frac{k}{n}$ for $1 \leq k \leq n$ and the right hand points of the interval as intermediate points.

Therefore $$\displaystyle\lim_{n \to \infty} {\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}}=\int_0^1 \frac{1}{1+x}dx =\ln(1+x)|_0^1=\ln(2)$$

P.S. One can reach the came conclusion by using the well known identity $$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}+...-\frac{1}{2n}$$ and the standard definition of the Euler -Mascheroni constant, but this approach is typically beyond calculus.

Just for fun, here is how you get the limit with the E-M constant $$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}+...-\frac{1}{2n} \\ =\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n}-2 (\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2n})\\ =\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n}- (\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n})\\ =\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n}- \ln(2n)- (\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n} -\ln(n))+\ln(2) $$ This converges to $\gamma-\gamma+\ln(2)$.

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  • $\begingroup$ Would you mind explaining why @N.Bar's answer won't be right? Because for me, that explanation seems reasonable. (So why the sum of infinite number of $0 ≠ 0$?) $\endgroup$
    – Yan Zhuang
    Jan 10, 2020 at 23:58
  • $\begingroup$ If each term is going to $0$, but the number of terms is going to infinity, the sum is NOT necessarily going to 0. Intuitively, the hige number of terms can compensate for the fact that each term is getting smaller. @YanZhuang $\endgroup$
    – N. S.
    Jan 11, 2020 at 0:00
  • $\begingroup$ Oh, so it is like one of the indeterminate forms introduced when learning L'hopitals'Rule: $0*\infty$ ? As a result, since we can't obtain a result back then, we can not say it equals to $0$ or $\infty$ here $\endgroup$
    – Yan Zhuang
    Jan 11, 2020 at 0:02
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    $\begingroup$ @YanZhuang Yes, if the number of terms goes to infinity is exactly like there. Limit laws work nicely for finitely many terms, but in that case the number of terms goes to infinity, and each term goes to 0... It is similar in idea to the $\infty \cdot 0$ case from L'H..... This is an issue which appears sometimes in Calculus, and can be subtle until you see it once or twice. $\endgroup$
    – N. S.
    Jan 11, 2020 at 0:04
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    $\begingroup$ @YanZhuang If you met Riemann Sums before, this is actually a nice, yet standard, problem about Riemann Sums. This is probably why your teacher put it on the exam. $\endgroup$
    – N. S.
    Jan 11, 2020 at 0:05
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Please Read: This is an incorrect answer. I will, however, leave it up to show a common mistake that can be made in this type of problem.

$$\lim_{n \to \infty}\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}$$

Remember that $$\lim_{n \to \infty}\frac{1}{n} = 0$$

From this theorem, and by applying limit properties, you would get

$$\lim_{n \to \infty}\frac{1}{n+1}+\lim_{n \to \infty}\frac{1}{n+2}+\lim_{n \to \infty}\frac{1}{n+3}+...+\lim_{n \to \infty}\frac{1}{n+n}$$

$$0+0+0+0+...$$ $$0$$

What's the mistake? $0+0+0...$ is an indeterminate form

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  • $\begingroup$ Thanks a lot! I thought too much when doing this problem. $\endgroup$
    – Yan Zhuang
    Jan 10, 2020 at 23:47
  • $\begingroup$ But if this question is when x -> 0 or any other number, is this question still doable? $\endgroup$
    – Yan Zhuang
    Jan 10, 2020 at 23:48
  • $\begingroup$ $0+0+0+..+0+...$ infinitely many times is NOT 0, it is an indeterminate form! $\endgroup$
    – N. S.
    Jan 10, 2020 at 23:52
  • $\begingroup$ So this will be like a divergent series or it will have a definite answer? As I will eventually add up to 1/n ? $\endgroup$
    – Yan Zhuang
    Jan 10, 2020 at 23:52
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    $\begingroup$ It is actually easy to fall into this trap, I think I did it once too many years ago... $\endgroup$
    – N. S.
    Jan 11, 2020 at 0:07
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If you know about harmonic numbers $$S_n=\sum_{i=1}^n \frac 1 {n+i}=H_{2 n}-H_n$$ For large values of $p$, we have $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^4}\right)$$ Apply it twice and simply to get $$S_n=\log (2)-\frac{1}{4 n}+\frac{1}{16 n^2}+O\left(\frac{1}{p^4}\right)$$ which shows the limit and how it is approached.

Moreover, this gives a quite good approximation of the partial sums. For example $$S_{10}=\frac{155685007}{232792560}\approx 0.6687714$$ while the above truncated expansion would give $$S_{10}\sim \log(2)-\frac{39}{1600}\approx 0.6687722$$

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