Question

What is the difference between
\[\lim_{n \rightarrow \infty} a_n=5\]
and
\[\sum_{n=1}^{\infty}a_n=5\]

Answer 1

\[\sum_{n=1}^{\infty}a_n=5 \iff \lim\limits_{k\to \infty} \sum_{n=1}^{k}a_n=5\]

Answer 2

yep :)

Answer 3

So the limit of a_n is the same as the limit of the sum of a_n?

How is that?

Answer 4

no that is different

Answer 5

I meant, the sum of a_n is the same as the limit of the sum of a_n

Answer 6

what he meant is the sum to infinity is the as taking the limit of the sum to some k
an n goes to infinity

Answer 7

yes but the sum to infinity

Answer 8

But in this case how is the limit of a_n the same as the sum of a_n?

Answer 9

this \[\lim_{n\to \infty} a_n\] is just the limit of a general term of a squence

Answer 10

hmm that is not true!

Answer 11

So, is the sum of a_n just the sum of all the limits of a_n
?

Answer 12

lets consider an example

Answer 13

the infinite sum of a_n's can be written as the limit of n sum's that what that says

Answer 14

That isn't what was said...
Take the following for example:
\[a_n=\frac{-1}{n(n-1)}=\frac{1}{n}-\frac{1}{n-1} \\ \lim_{n \rightarrow \infty}a_n=\lim_{n \rightarrow \infty}\frac{-1}{n(n-1)}=0 \\ \text{ but the \sum thingy ... } \\ \lim_{n \rightarrow \infty} \sum_{i=2}^{n} \frac{-1}{i(i-1)}=\sum_{n=2}^{\infty}(\frac{1}{n}-\frac{1}{n-1}) \\ =(\frac{1}{2}-\frac{1}{2-1})+(\frac{1}{3}-\frac{1}{3-1})+(\frac{1}{4}-\frac{1}{4-1})+ \cdots \\ +(\frac{1}{k}-\frac{1}{k-1})+(\frac{1}{k+1}+\frac{1}{k+1-1} + \cdots \\ =(\frac{1}{2}-1)+(\frac{1}{3}-\frac{1}{2})+(\frac{1}{4}-\frac{1}{3})+\cdots \\ + (\frac{1}{k}-\frac{1}{k-1})+(\frac{1}{k+1}+\frac{1}{k})+\cdots \\ \]
1/2's cancel
1/3's cancel
everything thing cancels except the -1

Answer 15

I understand what a sum is and I understand what a limit is, however I don't understand how in this particular instance they are both equal.

Answer 16

for my example we have
\[\lim_{n \rightarrow \infty}a_n=0 \\ \sum_{n=2}^{\infty}a_n=-1 \]

Answer 17

Yes, but in the question they are both equal to 5.

Answer 18

I didn't see that before
but one doesn't always imply the other

Answer 19

I wonder if we can think of a a_n such that both of what he says is true

Answer 20

Not possible. For the series to converge, the sequence must converge to 0

Answer 21

oh yeah true

Answer 22

\[\lim_{n \rightarrow \infty} a_n\ne 0 \implies \sum_{n=1}^{\infty}a_n \color{blue}{\text{ does not converge}}\]

Answer 23

http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries.aspx
I always love paul if you want to read any of this @akeith15 .

Answer 24

And that exact theorem mentioned by @ganeshie8 is also there.