Sum

Beeebooo A.

asked • 07/08/15

Calculate S3,S4, and S5 and then find the sum for the telescoping series S=?n=68(1n+1-1n+2) S3 = S4 = S5 = S =

Calculate S3,S4, and S5 and then find the sum for the telescoping series
S=∑(1/n+1−1/n+2),n=0..infinity
S3 =

S4 =

S5 =

S =

Stephanie M.

tutor
Beeebooo,
 
You've posted a few questions over the last couple days and I've noticed that you don't quite seem to be using the site correctly.
 
For your own sake, you should probably check over your questions before you post them to make sure they're written clearly. In this question, for example, you write "S=?n=68(1n+1-1n+2)," which doesn't really make any sense. Tutors who read that phrase will either have to try to guess what you mean (like Mark, below) and may wind up giving you incorrect answers because of that, or will have to ignore your question entirely.
 
In addition, it's usually not a good idea to respond to a tutor's attempt at answering your question with "still incorrect." That kind of comment is completely unhelpful---if you believe a tutor's answer is incorrect, give the tutor some idea of what you think is wrong with the answer. This opens a dialogue between student and tutor, so that together, the two can find the correct answer. It may be that the answer is correct and you've made a mistake, or it may be that the answer is incorrect because the tutor has made a mistake. Either way, saying someone is wrong but providing no extra information gets you nowhere.
 
Please keep in mind that WyzAnt's tutors do not have to answer these questions, and they don't get paid to do so. They do so voluntarily, on their own time, in order to help students in need. If you don't check the correctness of your question, you'll get incorrect answers. If you don't format your question in an understandable way, you'll get no answers. If you're rude or unhelpful to tutors, they likely won't try to answer your questions in the future.
 
Ultimately, tutors here enjoy explaining problems and concepts to students, and students benefit by having this incredible resource to turn to for help. However, you lose out on the benefit if you don't do your part---be clear in your questions, and help the tutors help you!
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07/08/15

Beeebooo A.

i am really sorry about that, but i thought i cant discuss the question with the tutor
and thank u becaus u informed me my mistakes. i am thankful to you for answers and helping me
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07/08/15

Mark M.

tutor
Stephanie: Thanks for your thoughtful comment.  I am waiting for                         "Beeebooo" to enlighten me on why my answer (which I                      think is correct) is "still incorrect".  Happy tutoring!  
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07/08/15

Beeebooo A.

actually i do not know why is incorrect, they just write the answer is wrong and i did not get point on it
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07/08/15

Mark M.

tutor
Who are "they"?  As Stephanie commented, maybe I answered the wrong question?  Can you restate the question clearly?  
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07/08/15

Beeebooo A.

Calculate S3,S4, and S5 and then find the sum for the telescoping series
S=∑((1/(n+1))−(1/(n+2)), n=6 to infinity
S3 =

S4 =

S5 =

S =
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07/08/15

Mark M.

tutor
Are you sure that the sum is from n = 6 to infinity?  
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07/08/15

Beeebooo A.

yes
under the symbol of sigma notation is n=6 and above infinity
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07/08/15

1 Expert Answer

By:

Mark M. answered • 07/08/15

Tutor
4.9 (883)

Math Tutor--High School/College levels
About this tutor ›
About this tutor ›
OK.  Here is my final attempt:
 
Σ[1/(n+1)-1/(n+2)] from n = 6 to ∞
 
 = [1/7-1/8] + [1/8-1/9] + [1/9 - 1/10] + ...
 
sum of first n terms 
 
     = [1/7-1/8]+[1/8-1/9]+[1/9-1/10]+...+ [1/(n+6) - 1/(n+7)] 
 
     = 1/7 - 1/(n+7)
 
S3 = sum of first 3 terms = 1/7 - 1/10 = 3/70 
S4 = sum of first 4 terms = 1/7 - 1/11 = 4/77
S5 = sum of first 5 terms = 1/7 - 1/12 = 5/84
 
The sum, S, of the entire series is given by:
 
      S = lim (Sn) as n →∞ = lim[1/7 - 1/(n+7)] as n→∞
 
                                           = 1/7 - 0 = 1/7
 
     
Upvote • 2 Downvote
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Beeebooo A.

still incorrect
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07/08/15

Beeebooo A.

still incorrect
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07/08/15

Beeebooo A.

thanks a lot
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07/08/15

Beeebooo A.

would u mind if i ask another question?
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07/08/15

Mark M.

tutor
OK.  Fire away.  
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07/08/15

Beeebooo A.

For each, find an appropriate function.
(a) Find a function f(k) such that
2−4+6−8+10=∑f(k), k=1 to 5
f(k) =

(b) Find a function g(k) such that
2−4+6−8+10=∑g(k), k=0 to 4
g(k) =
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07/08/15

Mark M.

tutor
2, 4, 6, 8, ... are multiples of 2
Use (-1)k or (-1)k+1 to make the signs alternate
 
(a).  Let f(k) = (-1)k+1(2k)
 
         Then Σ[(-1)k+1(2k)] from k = 1 to k = 5
 
                   = (-1)2(2)+(-1)3(4)+(-1)4(6)+(-1)5(8)+(-1)6(10)
 
                   = 2-4+6-8+10
 
(b).  Let g(k) = (-1)k(2k)
 
         Σ[(-1)k(2(k+1)) from k = 0 to k = 4 
 
         = (-1)0(2) + (-1)1(4) + (-1)2(6) + (-1)3(8) + (-1)4(10)
 
         = 2-4+6-8+10
 
          
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07/08/15

Beeebooo A.

thank u
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07/08/15

Mark M.

tutor
Correction:  g(k) = (-1)k(2(k+1)), not (-1)k(2k).
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07/08/15

Jacob L.

"sum of first n terms

= [1/7-1/8]+[1/8-1/9]+[1/9-1/10]+...+ [1/(n+6) - 1/(n+7)]

= 1/7 - 1/(n+7)"
 
This is an old thread, but it helped me out today.  Just one question, why does:
 
[1/(n+6) - 1/(n+7)]
 
become:
 
1/7 - 1/(n+7)
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09/29/15

Mark M.

tutor
The entire sum simplifies to 1/7 - 1/(n+7) because all the terms in between 1/7 and -1/(n+7) cancel out (a "telescoping" effect).  
 
Mark M  (Bayport, NY)  
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09/29/15

Jacob L.

Perfect.  Thanks for the reply.
 
May a thousand upvotes rain down upon you!
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09/29/15

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