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I am studying MIT OpenCourseware 18.01 Single Variable Calculus on my own and am stuck on a final exam question.

Evaluate the following limit: $$\lim\limits_{n\to \infty} \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}} \left(\frac{2}{n}\right)$$

$$=\lim\limits_{n\to \infty} \left(\frac{2}{n}\right) \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}} $$

We can do this using the Riemann sum, which states that if the interval [a,b] is divided into $n$ equal pieces of length, where $\Delta x = \frac{b-a}{n}$, then the sum of all the areas of the rectangle is $ \sum_{i = 1}^{n} f(x_{i-1}) \Delta x $. Also, in the limit as $n$ goes to infinity, the Riemann sum approaches the value of the definite integral:

$$\lim\limits_{n\to \infty} \sum_{i = 1}^{n} f(x_{i-1}) \Delta x =\int_a^b f(x)\,dx$$

In this case, $\Delta x = \frac{b-a}{n} = \frac{2}{n}$, and therefore, $b-a = 2$. Also $f(x_0) = \sqrt{1+\frac{2}{n}}$, $f(x_1) = \sqrt{1+\frac{4}{n}}$, $f(x_2) = \sqrt{1+\frac{6}{n}}$, and so on and so forth until we reach $n$.

How do we convert $$\lim\limits_{n\to \infty} \left(\frac{2}{n}\right) \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}} $$ to a definite integral?

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  • $\begingroup$ All you need to figure out is what $f(x)$ should be so that $f(x_i) = \sqrt{1+2i/n}$. But what is $x_i$? $\endgroup$ Nov 24 '19 at 23:30
  • $\begingroup$ As $i$ goes from $1$ to $n$, $x_i$ should partition the interval $(a,b)$ $\endgroup$ Nov 24 '19 at 23:33
  • $\begingroup$ Ah ok. $x_i = a + \Delta x \cdot i$ $\endgroup$
    – rplee
    Nov 24 '19 at 23:58
  • $\begingroup$ Please note that if a question of yours is closed, it is better to edit it than to ask a new version. Editing a closed question automatically puts it in the Reopen review queue, where people can vote to reopen it if they consider that it is now suitable. $\endgroup$
    – Arnaud D.
    Nov 25 '19 at 15:53
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Viewing uniform partition on $[0,2]$ with width $2/n$, then it is $\displaystyle\int_{0}^{2}\sqrt{1+x}dx$.

Viewing uniform partition on $[0,1]$ with width $1/n$, then it is $2\displaystyle\int_{0}^{1}\sqrt{1+2x}dx$.

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  • $\begingroup$ and those have the same value $\endgroup$ Nov 24 '19 at 23:42
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    $\begingroup$ Yes, OP should have noticed that it is just a matter of change of variable formula. $\endgroup$
    – user284331
    Nov 24 '19 at 23:44
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The answer is $2\int_0^{1} \sqrt {1+2x}dx =\frac 2 3(3^{3/2}-1)$.

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Just for your curiosity (without Riemann sum).

Assuming that you know about generalized harmonic numbers, you could approximate quite well the partial sums since $$ S_n= \frac{2}{n} \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}}=2 \sqrt{2}\,\frac{ H_{\frac{3 n}{2}}^{\left(-\frac{1}{2}\right)}-H_{\frac{n}{2}}^{\left(-\frac{1}{2}\right)} }{n^{3/2}}$$

Now, using the asymptotics $$H_p^{\left(-\frac{1}{2}\right)}=\frac{2 p^{3/2}}{3}+\frac{p^{1/2}}{2}+\zeta \left(-\frac{1}{2}\right)+\frac{1}{24p^{1/2}}-\frac{1}{1920p^{5/2}}+O\left( \frac{1}{p^{9/2}}\right)$$

$$S_n=\left(2 \sqrt{3}-\frac{2}{3}\right)+\frac{\sqrt{3}-1}{n}+\frac{\sqrt{3}-3}{18 n^2}+\frac{27-\sqrt{3}}{3240 n^4}+O\left(\frac{1}{n^6}\right)$$ which, for sure, shows the limit and also how it is approached.

Moreover, it is a quite good approximation. For example, computing $S_5\approx 2.9410395$ while the above truncated formula would give $\frac{4459499 \sqrt{3}-1768473}{2025000}\approx 2.9410399$.

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