-1

enter image description here

Would someone be so kind as to demonstrate the proof for me?

edit:

$$\sum_{n=v+1}^\infty a_n=a_{v+1}+a_{v+2}+\cdots$$ The $N^\text{th}$ partial sum is: $$\begin{align}\sum_{v+1}^{v+N} a_{n}&=a_{v+1}+a_{v+2}+\cdots+a_{v+N} \\&= (a_1+a_2+a_3+\cdots+a_{v+N})-(a_1+a_2+a_3+\cdots+a_v) \\&= \sum_{n=1}^{v+N} a+n-(a_1+a_2+\cdots+a_v)\end{align}$$

I do not understand the reasoning that follows through from the second to the third line. (LaTeX isn't working for some reason.

Edit: On a tangent, could someone advise me why my LaTeX code isn't working despite having it generated by Mathematica and an online editor?

guest
  • 387
  • You can assumed that the series $\sum_{n=1}^{\infty} a_n$ is convergent. Now you want to show that another series is convergent. Try applying the definition. – Mankind Mar 15 '15 at 12:02
  • You got downvoted because you didn't show any effort. What have you tried already? What do you already know about the problem? How do you think that you should solve the problem, but are maybe unable to do it, etc. – Pedro Mar 15 '15 at 12:33
  • I have the solution but I do not understand the proof. Was looking for a, perhaps, simpler proof. – guest Mar 15 '15 at 21:39

1 Answers1

1

An elementary proof: Consider the series $\sum b_n$ defined by $b_n=a_n$ if $n\leq \nu$ and $b_n=0$ if $n>\nu$. It is convergent to $\sum\limits_{n=1}^{\nu} a_n$, hence the series $\sum(a_n-b_n)$ is too, and $$\sum_{n=\nu+1}^{\infty} a_n=\sum_{n=1}^{\infty} a_n-\sum_{n=1}^{\infty} b_n=a -\sum_{n=1}^{\nu} a_n. $$

Bernard
  • 175,478
  • I think I'm seeing it after studying the proof further. I was confused badly by the notations. – guest Mar 16 '15 at 01:43