Finding the sum of 1/(n(n+1)) from n=1 to infinity

[martina1075]

Homework Statement

Good morning,
can someone help me solve this please?
Thank you in advance

Relevant Equations

As per picture

Cannot conclude the answer

 

[PeroK]

That's an interesting one. Does it say what you are supposed to do?

 

[martina1075]

That's an interesting one. Does it say what you are supposed to do?

It says to evaluate the sum. The answer should be 5. The problem is how he got to 5.

 

[PeroK]

It says to evaluate the sum. The answer should be 5. The problem is how he got to 5.

Have you tried with one or two values for $n$ to see what you get?

 

[berkeman]

I don't understand. Where is k in the sum? Shouldn't the sum be over n?

 

[Delta2]

if I tell you to evaluate $$\sum_{k=1}^n c$$ where c is some constant that does not depend on k(but might depend on other things like n for example) what will your answer be?

 

[PeroK]

I don't understand. Where is k in the sum? Shouldn't the sum be over n?

View attachment 251393

That's what I thought at first, hence post #2. But, then, I saw the light!

 

[osilmag]

There should be a table in your textbook for common arithmetic series. There is no way that the answer is 5.

 

[PeroK]

There should be a table in your textbook for common arithmetic series. There is no way that the answer is 5.

Read the question carefully!

 

[Mark44]

There should be a table in your textbook for common arithmetic series.

No table is needed.

There is no way that the answer is 5.

See Delta2's hint in post 6.

 

[osilmag]

Read the question carefully!

I did!

Five is if the index stops at one. It doesn't state that in the problem, but it sure tricked me.

 

[PeroK]

I did!

Five is if the index stops at one. It doesn't state that in the problem, but it sure tricked me.

The answer is $5$ for any value of $n$.

 

[Mark44]

Read the question carefully!

I did!

Read it even more carefully!

Five is if the index stops at one. It doesn't state that in the problem, but it sure tricked me.

If n = 2, we have
$$\sum_{k = 1}^2 5 \cdot \frac 1 2 = 5 \cdot \frac 1 2 + 5 \cdot \frac 1 2 = ?$$

What if n = 3? What if n = 4?

 

[osilmag]

Ok. I got it! Like I said it tricked me.

I had thought the index counted up from k to n.

 

[Mark44]

Ok. I got it! Like I said it tricked me.

I had thought the index counted up from k to n.

No, the index k counts up from 1 to n. Since the summand doesn't involve k, and so is constant, we're just adding $5 \cdot \frac 1 n$ n times.

 

[PeroK]

No, the index k counts up from 1 to n. Since the summand doesn't involve k, and so is constant, we're just adding $5 \cdot \frac 1 n$ n times.

... which is essentially the definition of $n \times \frac 5 n$.

 

[HallsofIvy]

... which is essentially the definition of $n \times \frac 5 n$.

But is not the usual definition of "5"!