help with summation of finite series

**Kuma** · September 23rd 2011, 04:17 PM

I want to know how to calculate this finite sum

sum of i=1 to 52 of (i^2/3)*1/60

what method can I use for this?

**FernandoRevilla** · September 23rd 2011, 04:49 PM

Originally Posted by Kuma

I want to know how to calculate this finite sum sum of i=1 to 52 of (i^2/3)*1/60 what method can I use for this?

Hint Use the well known formula $\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$

**Soroban** · September 23rd 2011, 07:25 PM

Hello, Kuma!

I want to know how to calculate this finite sum

. . sum of i=1 to 52 of (i^2/3)*1/60

What method can I use for this?

Is that: . $\frac{1}{60}\sum^{52}_{i=1}i^{\frac{2}{3}}$

**Kuma** · September 23rd 2011, 08:53 PM

Gotcha, thanks guys!

**Kuma** · September 24th 2011, 08:36 AM

hmm seems I'm getting an incorrect answer. Can I split i^2/3 as (i^2)^1/3 and then use the i^2 formula? I don't know if that is the same. I figure my wrong answer is because I used the i^2 formula, plugged in n as 52 and cubed rooted that number, then divided by 60.

**chisigma** · September 24th 2011, 09:40 AM

Originally Posted by Kuma

I want to know how to calculate this finite sum

sum of i=1 to 52 of (i^2/3)*1/60

what method can I use for this?

If the sum to be computed is...

$S= \sum_{i=1}^{52} i^{\frac{2}{3}$ (1)

.... then, if You transform the (1) in

$S=1+\sum_{i=2}^{n} i^{\frac{2}{3}} = 1 + \sum_{k=1}^{n-1}(1+k)^{\frac{2}{3}}$ (2)

..., apart the 'direct computation' of course, You can consider to use of the 'Euler-McLaurin formula'...

$\sum_{k=1}^{n-1} f(k)= \int_{0}^{n} f(t)\ dt -\frac{1}{2}\ \{f(0)+f(n)\} + \frac{1}{2}\ \{f^{'}(n)-f^{'}(0)\} -$

$-\frac{1}{720}\ \{f^{(3)}(n)- f^{(3)}(0)\} + \frac{1}{30240}\ \{f^{(5)}(n)-f^{(5)}(0)\} -$

$- \frac{1}{1209600}\ \{f^{(7)}(n)-f^{(7)}(0)\} +...$

$...+ \frac{B_{2k}}{(2k)!}\ \{f^{(2k-1)}(n)-f^{(2k-1)} (0)\} +...$ (1)

... where the $B_{2k}$ are the 'Bernoulli's numbers'. In Your case is $f(t)= (1+t)^{\frac{2}{3}}$ and $n=52$ . The computation of the $f^{(2k-1)}(*)$ is left to You. It is difficult to extablish for what n the Euler-McLaurin formula [which of course is 'approximative'...] is 'competitive' with the direct computation...

Kind regards

$\chi$ $\sigma$

**Kuma** · September 24th 2011, 04:38 PM

I still don't get it...do I have to use the euler formula for this? Wlfram alpha gave me the answer but my manual calculation is incorrect.
help? This is what I did

1/60 [(n)(n+1)(2n+1)÷6]^1/3 and plug n in as 52.

**FernandoRevilla** · September 24th 2011, 09:28 PM

Originally Posted by Kuma

I want to know how to calculate this finite sum
sum of i=1 to 52 of (i^2/3)*1/60 what method can I use for this?

If you don't use $\LaTeX$ , be careful with brackets

sum of i=1 to 52 of (i^2/3)*1/60 means $\sum_{i=1}^{52}\frac{i^2}{3}\cdot \frac{1}{60}$

sum of i=1 to 52 of ( i^(2/3) )*1/60 means $\sum_{i=1}^{52}i^{\frac{2}{3}}\cdot \frac{1}{60}$

**Kuma** · September 24th 2011, 10:00 PM

Yes it is the second one. I can't use the i^2 formula on it. How do I solve this?? I've been trying to figure out if fits into some sort of series like power series but all of those are infinite that involve convergence/divergence. What should I do?!

**chisigma** · September 24th 2011, 11:18 PM

Originally Posted by Kuma

Yes it is the second one. I can't use the i^2 formula on it. How do I solve this?? I've been trying to figure out if fits into some sort of series like power series but all of those are infinite that involve convergence/divergence. What should I do?!

Why are You so discouraged?... Your task is the computation of...

$S=\sum_{i=1}^{52} i^{\frac{2}{3}}$ (1)

... You have two possibilities...

a) if a computer is available for You the 'brute force sum' gives as result...

$S_{52,BF}= 441.492660573...$

b) if only a [scientific...] calculator and limited time is available for You, the Euler-McLaurin formula, that I described few post ago, may be useful...

Now we write the value of $S_{n, EMc}$ obtained with the Euler-McLaurin formula 'breaked up' to the third term...

$S_{n,EMc}= 1 + \int_{0}^{n} f(t)\ dt -\frac{1}{2}\ \{f(0)+f(n)\} +$

$+ \frac{1}{2}\ \{f^{'}(n)-f^{'}(0)\} - \frac{1}{720}\ \{f^{'''}(n)-f^{'''}(0)\}$ (2)

... where is $f(t)=(1+t)^{\frac{2}{3}}$ . If You insert in (2) n=52 You obtain...

$S_{52,EMc}= 441.333255853...$

If more precision is requested, You can use more terms of the Euler-McLaurin formula. May be interesting to compare the 'brute force sum' and the 'Euler-McLaurin third degree sum' for greater values of n...

$S_{100,BF}= 1303.28975964...\ ,\ S_{100,EMc}= 1303.10747032...$

$S_{1000,BF}= 60049.8503587...\ ,\ S_{1000,EMc}= 60049.6170634...$

$S_{10000,BF}= 2.78518522699... 10^{6}\ ,\ S_{10000,EMc}= 2.78518496989... 10^{6}$

Kind regards

$\chi$ $\sigma$

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