I was watching some lessons on Khan Acacemy and there was a section on partial sums https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-1/v/term-from-partial-sum
It was stated that given $S=\sum_{n=1}^{\infty}a_{n} $,
The partial sum defined as $S_{n}=\frac{n^{2}-3}{n^{3}+4}$
Will give the the sum of 1 through n terms of $a_{n}$
$S_{n}=a_{1}+a_{2}+…+a_{n-1}+a_{n}=\frac{n^{2}-3}{n^{3}+4}$
$\sum_{n=1}^{6}a_{n}=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=S_{6}$
$S_{6}= \frac{6^2-3}{6^3+4}=\frac{33}{220} = \sum_{n=1}^{6}a_{n}=\frac{33}{220}$
The problem I have is that when I try to test these methods out on another problem, we reach a false conclusion.
Given the following problem:
The nth partial sum of the series $\sum_{n=1}^{\infty}a_{n}$ is given by $S_{n} = \frac{n+1}{n+10}$
Write a rule for $a_{n}$
The approach taken is $a_{n} = S_{n} – S_{n-1} = \frac{9}{n²+19n+90}$
So if $\sum_{1}^{\infty}\frac{9}{n^2+19n+90}$, then I expected
$\sum_{1}^{6}a_n = S_{6}$
but…
$\sum_{1}^{6}\frac{9}{n^2+19n+90} ≠ \frac{(6)+1}{(6)+10}$
$.3375 ≠ .4375$
I've reduced this down to.
1) The false conclusion reached via testing is correct and the given partial sum rule rule is wrong.
2) I misunderstood what was asserted about comparing $S_{n}$ to $\sum_{1}^{n}$ and I'm testing a broken assumption.
3) The partial sum rule is correct and I messed up a step somewhere
What is wrong here, and why?
Best Answer
Note that $a_n=S_n-S_{n-1}$ only for $n\ge2$ since $S_p$ is defined only for $p\ge1$. Further, $a_1=S_1=2/11$. This$$a_n=\begin{cases}\frac9{(n+9)(n+10)},&n>1\\\frac2{11},&n=1\end{cases}$$and$$S_6=2/11+\sum_{n=2}^6\frac9{n^2+19n+90}=\frac7{16}$$which resolves the disparity.