Convince me: limit of sum of a constant is infinity

1.7k Views Asked by At

So I have a problem and have simplified the part I am confused about below.

If $\sum_{m=1}^{\infty }c < \infty$ and $0 \leq c \leq 1$, then $lim_{n\rightarrow \infty} \sum_{m=n}^{\infty }c= 0$ which implies $c=0$.

My general intuition says that because the sum of infinitely many non-negative c's is less than infinity, than $c=0$ because the sum of an infinitely many positive numbers will always be infinity.

The limit is where I am confused. I feel like the limit will always be $0$ even if $c>0$. It also feels like the limit is not necessary to show $c=0$.

3

There are 3 best solutions below

1
On BEST ANSWER

If $c>0$ then $\sum_{i=1}^{\infty }c= \lim_{n\to \infty } \sum_{i=1}^{n}c=\lim_{n\to \infty }nc =\infty $

If c=$0$ then $\sum_{i=1}^{\infty }c=0 $

If $c<0 $then $\sum_{i=1}^{\infty }c =-\infty $

1
On

Suppose that every day I go to the bank and deposit the same amount of money: $c$ dollars.

I want to buy a gold chain that costs $M$ dollars. Eventually, if I am diligent and keep depositing $c$ dollars every day, I will have enough to buy my gold chain, right? No matter how much $M$ is.

The only way this doesn't work is if the amount of dollars I am depositing every day is $0$.

0
On

Since your sum $$\sum _1^\infty C <\infty$$

We may apply the divergence test to conclude that $$\lim _ {n\to \infty}C=0.$$

Since C is a constant we have $$\lim _{n\to \infty }C =C.$$

Thus $C=0.$