How to find the sum: $1^{\frac{1}{3}}+2^{\frac{1}{3}}+3^{\frac{1}{3}}+ . . . +(50)^{\frac{1}{3}}$ Can some one help me to find the sum of the following expression?
$$1^{\frac{1}{3}}+2^{\frac{1}{3}}+3^{\frac{1}{3}}+ . . .  +(50)^{\frac{1}{3}}$$
 A: 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 $139.7179...$
A: 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.
A: There is no closed form for it, however it can be approximated by $\displaystyle\int_1^{50}\sqrt[3]x~dx=\bigg[\dfrac34x\sqrt[3]x\bigg]_1^{50}\approx$ $\approx137.4$
A: 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 $
A: 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 :
$$S_{50} = \zeta\biggl(-\frac13\biggr) - \zeta\biggl(-\frac13, 51\biggr) \approx 140$$
A: 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.
A: You can get as much decimal expansion as you please at WA.
139.7179
A: $$6+3 \sqrt[3]{2}+3\ 2^{2/3}+3 \sqrt[3]{3}+2^{2/3} \sqrt[3]{3}+3^{2/3}+\sqrt[3]{2} 3^{2/3}+3 \sqrt[3]{5}+2^{2/3} \sqrt[3]{5}+3^{2/3} \sqrt[3]{5}+5^{2/3}+\sqrt[3]{2} 5^{2/3}+3 \sqrt[3]{6}+6^{2/3}+\sqrt[3]{7}+2^{2/3} \sqrt[3]{7}+7^{2/3}+\sqrt[3]{10}+\sqrt[3]{11}+2^{2/3} \sqrt[3]{11}+\sqrt[3]{13}+\sqrt[3]{14}+\sqrt[3]{15}+\sqrt[3]{17}+\sqrt[3]{19}+\sqrt[3]{21}+\sqrt[3]{22}+\sqrt[3]{23}+\sqrt[3]{26}+\sqrt[3]{29}+\sqrt[3]{30}+\sqrt[3]{31}+\sqrt[3]{33}+\sqrt[3]{34}+\sqrt[3]{35}+\sqrt[3]{37}+\sqrt[3]{38}+\sqrt[3]{39}+\sqrt[3]{41}+\sqrt[3]{42}+\sqrt[3]{43}+\sqrt[3]{46}+\sqrt[3]{47}$$
