# Sum question

• Feb 1st 2010, 05:20 AM
Henryt999
Sum, not solved yet...
I´m trying to learn something about matematical induction. This chapter starts with sums (Sigma)

There is this easy problem I do not understand.
It says calculate the Sigma: from k=1;to n. where \$\displaystyle (1/k-1/(k+1))\$

But when I do it I get:
starts at k=1 so I get: \$\displaystyle 1/1-1/(1+1) = 1/2\$

Now I try to evaluate the last term:

\$\displaystyle 1/n-1/(n+1)\$ /multiply by n(n+1)
= \$\displaystyle (n+1)/n(n+1) -n/(n(n+1)) \$
that gives me \$\displaystyle 1/(n^2+n)\$

so I get \$\displaystyle 1/2 +1/(n^2+n)\$
Now to evaluate the sum I take the (first term +last term)multiply by n and devide by 2.

That gives me \$\displaystyle (n(1/2+1(n^2+n))/2 \$= \$\displaystyle n/4+1/(2(n+1))\$

clearly Iám doing someting very very stupid...anyone got an idéa whats wrong here?
• Feb 1st 2010, 05:34 AM
HallsofIvy
Quote:

Originally Posted by Henryt999
I´m trying to learn something about matematical induction. This chapter starts with sums (Sigma)

There is this easy problem I do not understand.
It says calculate the Sigma: from k=1;to n. where \$\displaystyle (1/k-1/(k+1))\$

But when I do it I get:
starts at k=1 so I get: \$\displaystyle 1/1-1/(1+1) = 1/2\$

Now I try to evaluate the last term:

\$\displaystyle 1/n-1/(n+1)\$ /multiply by n(n+1)
= \$\displaystyle (n+1)/n(n+1) -n/(n(n+1)) \$
that gives me \$\displaystyle 1/(n^2+n)\$

Okay, those are correct.

Quote:

so I get \$\displaystyle 1/2 +1/(n^2+n)\$
Now to evaluate the sum I take the (first term +last term)multiply by n and devide by 2.
Why? What reason do you have to think that would work? For example, if I want to sum 2+ 8+ 9+ 11, the first number is 2, the last is 11, and there are 4 numbers. Their average, (2+11)/2, multiplied by 4 is 25. but the sum is actually 30.

Actually, I do know why you did that- for an arithmetic series that is true, and a very useful fact. But it is true only for arithmetic series, where each term is the previous term plus a fixed number. For example, 2, 8, 14, 20, ... has each number the previous number plus 6. In the example I gave, 2+ 8+ 9+ 11, the second number is 2+ 6, but the third number is 8+ 1 and the fourth number is 9+ 2. Not a constant difference.

For your series, the first number is, as you say 1/2. The second number is 1/2- 1/3= 3/6- 2/6= 1/6 and the third number is 1/3- 1/4= 4/12- 3/12= 1/12.
1/3= 1/2+ (-1/6) but 1/12= 1/3+ (-3/12). NOT an arithmetic series.

Quote:

That gives me \$\displaystyle (n(1/2+1(n^2+n))/2 \$= \$\displaystyle n/4+1/(2(n+1))\$

clearly Iám doing someting very very stupid...anyone got an idéa whats wrong here?
Well, not very, very; perhaps too many "very"s!(Giggle)
• Feb 1st 2010, 05:45 AM
Henryt999
Ahaa
Thanks!! With your help now have one less "very". Ill try to make those verys => lim: "very stup.." =>0 (Rock)
• Feb 1st 2010, 07:38 AM
Henryt999
I´m sorry I still don´t get how to get the answer
So I have the same issue. I posted 1.5hours ago.
Thank you hallsofIvy but I don´t grasp how to continue.

Sigma from k=1 to n \$\displaystyle (1/k-1/(k+1))\$

How on earth do I get that to be \$\displaystyle 1-1/(1+n)\$?

I tried this \$\displaystyle 1/2+1/6+1/12 +1/(n-1) -1/n\$. Then I try to get every term that includes an n to the same denominator. so I get \$\displaystyle n/n(n-1) -(n-1)/n(n-1)\$

so I get \$\displaystyle (n-n+1)/n(n-1) = 1/(n^2-n)\$ and ofcourse that is almost very very not right.

So well that didn´t work.

Well I tried the other way \$\displaystyle 1/2+1/6....+1/n-1/(n+1)\$ that gives me \$\displaystyle 1/(n^2+n)\$

And that doesn´t help....I really don´t need a complete solution maybe a hint would work wonders....
• Feb 1st 2010, 12:23 PM
girdav
You can try to write the sum for some values of \$\displaystyle n\$, for example until \$\displaystyle n=5\$.
You will understand how to calculate that for any \$\displaystyle n\$.
• Feb 1st 2010, 04:24 PM
wonderboy1953
Induction
Did you let \$\displaystyle k = k + 1\$?