# Convince me: limit of sum of a constant is infinity

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 $$c=0$$.

• I think you are missing something in your question. Are you really dealing with a constant. Or are you dealing (for example when dealing with integration) a function of some sort? – Q the Platypus Nov 15 '18 at 5:13
• I think you are confused with "fixing" the value of $c$ first since you are talking about constants. If $c = 0$ has been fixed beforehand, of course the sum is zero but if $c>0$ is chosen instead, then you will tend to infinity by taking the limit of partial sum of the series. You can see clearly the limit tends to infinity simply by looking at the definition of limit. – Evan William Chandra Nov 15 '18 at 5:16
• I'm not actually dealing with a constant. I have a sequence {X_n} in which a term in the sequence occurs infinitely often, say {X_i}. I then have a function for each term f(X_j) in which the function 0<=f(X_j)<=1. for all X_j. The question is asking us to show that f(X_i), which occurs infinitely often, must be equal to zero if the sum of all f(X_j) is less than infinity. – kpr62 Nov 15 '18 at 5:22
• The solution uses a limit, and I don't understand why. – kpr62 Nov 15 '18 at 5:24
• @kpr The sum of your subsequence of infinitely occurring $f(X_i)$s is a lower bound for the sum of the whole sequence. You're right that the problem reduces to what you stated in your question. – Alexander Gruber Nov 15 '18 at 5:25

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 $$\sum_{i=1}^{\infty }c =-\infty$$

• Add this with @Alexander Gruber's comment, and I am convinced. – kpr62 Nov 15 '18 at 5:28

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 $$0$$.

• it is a good example/explination. But I think you should explain how your example connects to the definition of having a limit equal infinity. – Q the Platypus Nov 15 '18 at 5:18

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 $$C=0.$$