How can I find this sum?

KFS

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How do I find the sum of 1/(1+k^2) from k=1 to 1000? The book didn`t provide any information of how to solve rational sums, nor formulas for greater powers than 1. I have no idea how to do this. Thank you.
 
How do I find the sum of 1/(1+k^2) from k=1 to 1000? The book didn`t provide any information of how to solve rational sums, nor formulas for greater powers than 1. I have no idea how to do this.
It would extremely helpful to know the context in which this question occurs. Are you expected to simply find the sum? If that is the case HERE IT IS.
On other hand if you want more, we need to know more context.
 
What DOES the book show for series similar to this? We need to know what techniques are available to you. What formulas were given? What methods are used in examples or in other exercises?
 
This is what the book gives.
 

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My first thought when I saw the problem was that if it had been something like SUM[1/(k^2 - 1)], it would telescope. But since 1 + k^2 doesn't factor, that won't work. So I was sort of hoping it would be a copying error!

I can't think of any other methods at your level. In general, series are difficult problems.

Again, please quote the actual wording of the problem (and any instructions for the set of problems), so we can be sure what is expected.
 
My first thought when I saw the problem was that if it had been something like SUM[1/(k^2 - 1)], it would telescope. But since 1 + k^2 doesn't factor, that won't work. So I was sort of hoping it would be a copying error!

I can't think of any other methods at your level. In general, series are difficult problems.

Again, please quote the actual wording of the problem (and any instructions for the set of problems), so we can be sure what is expected.
It says: Show that the sum from k=1 to 1000 of 1/(1+k^2) is smaller or equal to 1000.
 
It says: Show that the sum from k=1 to 1000 of 1/(1+k^2) is smaller or equal to 1000.
OK so you don't actually have to find the sum you just need to show it is less than or equal to 1000. That makes it a bit easier. (BTW are you sure the 1000 is not a typo??)

Note that \(\displaystyle \frac{1}{1+k^2}\leq\frac{1}{k^2}\).

So the sum you are after is less than or equal to the sum of \(\displaystyle \frac{1}{k^2}\) for the same values of k. (Property 5 on your list.)

Can you find this sum?

Another (better) thought (assuming the 1000 is not a typo):
Each term in the series is less than 1, so the sum of the series is going to be less than (but not equal to) 1000 x 1. Done.
 
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It says: Show that the sum from k=1 to 1000 of 1/(1+k^2) is smaller or equal to 1000.
Please let this be a lesson; if you continue to post, please always give us the complete question. It saves time.
\(\displaystyle \text{for all }k\;,1+k^2\ge 1\) therefore
\(\displaystyle \frac{1}{1+k^2}\le 1\) that give us
\(\displaystyle \sum\limits_{k = 1}^{1000} {\frac{1}{{1 + {k^2}}} \le \sum\limits_{k = 1}^{1000} 1 = 1000}\)
 
It says: Show that the sum from k=1 to 1000 of 1/(1+k^2) is smaller or equal to 1000.

Now I see that they want you to use the 5th property of summation in the page you showed! Since you didn't mention an inequality, I didn't consider that as being relevant.

Now let's think about why you left out the most important part of the problem.

I imagine that you assumed that if they ask you to show that the sum is less than a certain number, they expect you to find the sum in order to do that. In most school math, you are only given problems that can be solved that directly. But if you are in calculus and studying series, you are old enough to know the truth: Most equations can't be solved, and most problems can't be answered exactly! You first learn only the problems that can be solved easily, and only later learn about real life, if ever.

Such naivete is the main reason we ask everyone to show the entire problem they are working on, not just the piece they think is important: You don't know yet what matters, or how wrong your initial thinking may be! That's true on any topic that is new to you.
 
Thanks everyone for your answers, it helped. (1000 is not a typo, it is correct.)
 
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