Why does s_{n-1} start from a_1 and not a_0?

[flyingpig]

Homework Statement

$$Prove\; that\;if\;\sum_{n=1}^{\infty} a_n \;converges,\;then \lim_{n\to\infty}a_n = 0$$

Book solution

$$s_n= a_1 + a_2 +...+a_n$$

$$s_{n-1}= a_1 + a_2 +...+a_{n-1}$$

$$a_n=s_n-s_{n-1}$$

Then they did a few limits, and proved that the difference is 0. BUt that is not my question.

My question is this part
$$s_{n-1}= a_1 + a_2 +...+a_{n-1}$$

If it is n - 1, why are they starting from a₁? Shouldn't it be a_0/sub]?

 

[gb7nash]

No. If we're assuming that the first term in the sum is $$a_{1}$$, the (n-1)th partial sum is defined to be $$s_{n-1}= a_1 + a_2 +...+a_{n-1}$$, i.e. it's the sum of the first term, the second term, ... , and the (n-1)th term.

 

[flyingpig]

Yeah exactly so it should be a₀

 

[gb7nash]

In your problem, the first term will always be $$a_{1}$$. The partial sum that we choose won't affect the first term. We could have $$s_{n}$$, $$s_{n-1}$$, $$s_{n+3}$$, but in each case the first term will always be $$a_{1}$$

If you're still not convinced, take a look at:

http://mathworld.wolfram.com/PartialSum.html

However, if the series defined is given as $$\sum_{n=0}^{\infty} a_n$$, then you would be right. The first term of all partial sums would start at $$a_{0}$$. It just depends on the first term that's defined.