Partial Sum Not Equal to Sum Of n terms

Question

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?

Answer 1

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.

Answer 2

$$S_n:=\sum_{k=1}^n a_k$$

and

$$S_n-S_{n-1}=a_n$$

only hold for $n>1$. For $n=1$, we simply have

$$S_1=a_1.$$

With your second example,

$$a_n=\frac9{(n+9)(n+10)}$$ but $$a_1=\frac{1+1}{1+10}\ne\frac9{(1+9)(1+10)}.$$

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Said differently,

$$S_n=\frac{n+1}{n+10}$$ cannot hold with $n=0$, because the sum is void and

$$S_0=0.$$