How to find the sum: $1^{\frac{1}{3}}+2^{\frac{1}{3}}+3^{\frac{1}{3}}+ . . . +(50)^{\frac{1}{3}}$

Question

Can some one help me to find the sum of the following expression?

$$1^{\frac{1}{3}}+2^{\frac{1}{3}}+3^{\frac{1}{3}}+ . . . +(50)^{\frac{1}{3}}$$

Answer 1

Of course a computer can evaluate it. You can also view this and similar sums in terms of a Riemann sum: $$50^{1 \over 3}*50*\bigg[{1 \over 50} \bigg({1 \over 50}\bigg)^{1/3} + {1 \over 50}\bigg({2 \over 50}\bigg)^{1 \over 3} + ... + {1 \over 50}\bigg({50 \over 50}\bigg)^{1 \over 3}\bigg]$$ This is approximately $$50^{1 \over 3}*50\int_0^1 x^{1 \over 3}\,dx$$ $$=50^{1 \over 3}*50*{3 \over 4}$$ $$= 138.1512...$$ This is reasonably close to the true value $139.7179...$

Answer 2

Because $f(x)=x^{1/3}$ is increasing on $(0,\infty)$, you can get bounds on the sum from

$$\int_0^{50}x^{1/3}dx\lt\sum_{k=1}^{50}k^{1/3}\lt\int_1^{51}x^{1/3}dx$$

Evaluating the integrals gives

$$138.151\lt\sum_{k=1}^{50}k^{1/3}\lt141.09744$$

You can get a better pair of bounds from

$$1+\int_1^{50}x^{1/3}dx\lt1+\sum_{k=2}^{50}k^{1/3}\lt1+\int_{2}^{51}x^{1/3}dx$$

which gives

$$138.4\lt\sum_{k=1}^{50}k^{1/3}\lt140.96$$

You can also try the approximation

$$\sum_{k=1}^{50}k^{1/3}\approx1+\int_{1.5}^{50.5}x^{1/3}dx\approx139.708$$

but it's hard to tell how many digits of that answer you can trust.

Answer 3

There is no closed form for it, however it can be approximated by $\displaystyle\int_1^{50}\sqrt[3]x~dx=\bigg[\dfrac34x\sqrt[3]x\bigg]_1^{50}\approx$ $\approx137.4$

Answer 4

You can bound the sum for any upper limit b.

$\frac{3 b^{4/3}}{4}<\sum _{n=1}^b \sqrt[3]{n}< \frac{3}{4} \ \left(-1+\sqrt[3]{1+b}+b \sqrt[3]{1+b}\right)$

Plugging in b = 50 you would get

$ 138.151<\sum _{n=1}^{50} \sqrt[3]{n}<141.097 $

Answer 5

If you write : $$S_n = \sum_{i = 1}^n i^{1/3}$$ then notice that you can create the following difference equation : $$S_n = S_{n-1} + n^{1/3}$$

This has no "nice" solution but only in terms of the Riemann and Hurwitz zeta functions:

$$S_n = \zeta\biggl(-\frac13\biggr) - \zeta\biggl(-\frac13, n+1\biggr)$$

Using WA, you find that : $$S_{50} = \zeta\biggl(-\frac13\biggr) - \zeta\biggl(-\frac13, 51\biggr) \approx 140$$

Answer 6

By the Euler-Maclaurin formula the sum $$ S = (1/2) 1^{1/3} + (2^{1/3} + 3^{1/3} + \cdots + 49^{1/3}) + (1/2) 50^{1/3} $$ can be approximated by the sum $$ I = \int_1^{50} x^{1/3} \: dx $$ and therefore your sum can be approximated by $$ I + (1/2) 1^{1/3} + (1/2) 50^{1/3} $$. Doing the integral, this is $$ (3/4) (50^{4/3} - 1^{4/3}) + (1/2) 1^{1/3} + (1/2) 50^{1/3} $$ or about 139.743. The true value is about 139.720, so this is already quite accurate.

This is actually nothing but the trapezoid rule, which works well because the function $f(x) = x^{1/3}$ is so smooth.

To get a more accurate formula for this sum, you can use the asymptotic form of the Euler-Maclaurin formula (under "asymptotic expansion of sums" in the Wikipedia article); I leave the details to someone else.

Answer 7

You can get as much decimal expansion as you please at WA.

139.7179

Answer 8

$$6+3 \sqrt[3]{2}+3\ 2^{2/3}+3 \sqrt[3]{3}+2^{2/3} \sqrt[3]{3}+3^{2/3}+\sqrt[3]{2} 3^{2/3}+3 \sqrt[3]{5}+2^{2/3} \sqrt[3]{5}+3^{2/3} \sqrt[3]{5}+5^{2/3}+\sqrt[3]{2} 5^{2/3}+3 \sqrt[3]{6}+6^{2/3}+\sqrt[3]{7}+2^{2/3} \sqrt[3]{7}+7^{2/3}+\sqrt[3]{10}+\sqrt[3]{11}+2^{2/3} \sqrt[3]{11}+\sqrt[3]{13}+\sqrt[3]{14}+\sqrt[3]{15}+\sqrt[3]{17}+\sqrt[3]{19}+\sqrt[3]{21}+\sqrt[3]{22}+\sqrt[3]{23}+\sqrt[3]{26}+\sqrt[3]{29}+\sqrt[3]{30}+\sqrt[3]{31}+\sqrt[3]{33}+\sqrt[3]{34}+\sqrt[3]{35}+\sqrt[3]{37}+\sqrt[3]{38}+\sqrt[3]{39}+\sqrt[3]{41}+\sqrt[3]{42}+\sqrt[3]{43}+\sqrt[3]{46}+\sqrt[3]{47}$$