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December 26th, 2017, 03:14 PM   #11
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I am late to this thread but

$\displaystyle \sum_{j=k}^n \text { expression in j and n }$

means to loop through, incrementing j by 1 each time, starting with j = k and ending with j = n, to calculate the value of the expression in each iteration of the loop, and to add that value to a cumulative sum.

$\displaystyle \sum_{j=2}^5 nj^2$ is computed as follows

Initialize sum = 0

First iteration j = 2 so

$nj^2 = 5 * 2^2 = 20 \text { and } 0 + 20 = 20.$

Second iteration j = 3 so

$nj^2 = 5 * 3^2 = 45 \text { and } 45 + 20 = 65.$

Third iteration j = 4 so

$nj^2 = 5 * 4^2 = 80 \text { and } 80 + 65 = 145. $

Last iteration j = 5 so

$nj^2 = 5 * 5^2 = 125 \text { and } 125 + 145 = 270.$

Summation notation is a highly abbreviated way to explain a procedure. Nothing fancy about it.
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December 28th, 2017, 03:32 PM   #12
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Quote:
Originally Posted by R_W_B View Post
. . . . . < cut > . . . . .
And lastly when I attempted to plug
$\sum \limits_{n=0}^\infty~\dfrac{(2n)!}{2^{2n}(n!)^2(2n +1)}x^{2n+1}$

into an online equation calculator, it gave this:
$[ \sum \limits_{n=0}^\infty~\dfrac{(2n)!}{2^{2n}(n!)^2(2n +1)} ]x^{2n+1}$

but (to me) that would give an errant answer for any iterations of n = greater than 1, since it seems to separate the x^(2n+1) from the Sum. But maybe that calculator knows more than me (?) at this point.
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Originally Posted by JeffM1 View Post
I am late to this thread but . . . . . .
Thanks for the reply, just now getting back on. Yes I understand all that now; my only confusion at this point is (shown above) when I plug into some equation calculators and hit enter, they put the parenthesis around the Sigma but not including the x^(2n+1) and this seems to me to exlude it from all iterations. Just a curiosity of why they do that or would that be wrong?

Last edited by skipjack; December 28th, 2017 at 04:02 PM.
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December 28th, 2017, 04:07 PM   #13
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Some calculators may be more flexible than others as to how a summation is entered.
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