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0
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avatar+1124 

Hi friends,

 

I have this sum..\(\sum_{y=3}^{10} {1 \over{y-2}}-\sum_{y=3}^{10} {1 \over{y-1 }}\)

What I did was:

For the first summation:

T1=1; T2=1/2...up to T10=1/8 = 761/280

For the second Summation

T1=1/2, T2=1/3...up to T10=1/9 = 1,828

 

Then just subtracting 1,828 from 761/280, and getting 0,89 as final answer..

 

Was my approach correct at all?..Thank you all for confirming/advising..I do appreciate.

 Apr 2, 2023
 #1
avatar+33616 
+1

It's perhaps a little more elegant the following way:

 Apr 2, 2023
 #2
avatar+1124 
0

HI Alan,

 

this is beautiful, thank you vwery much.

 

I do however have 2 questions please...

 

I would have mistakenly re-written the first term as \(1,5+\sum_{k=3}^{10}{1 \over k}\)

and then from here \(\sum_{k=1}^{10}{1 \over k}-1,5\)

This would have been completely wrong, and would not lead me in any way towards the end?

I have never seen the upper limit changed, as you have done?...Would ytou kindly just explain that to me, and just one thing more:  The second term is re-written as \({1 \over k}-1+{1 \over 9}\)

 

I'm looking hard here but fail to see how you got to that?...Thank you for your time..

juriemagic  Apr 2, 2023
 #4
avatar+33616 
0

The upper limit can be written as y = 10, though the "y" is usually dropped for the upper limit.

 

With y - 2 = k what you do is ask what is the value of k when y = 3  (k = 3 - 2 = 1), and what is the value of k when y = 10 (k = 10 - 2 = 8).

 

\(\sum_{k=2}^{k=9}\frac{1}{k}=\sum_{k=1}^{k=9}\frac{1}{k}-\frac{1}{1} =\sum_{k=1}^{k=8}\frac{1}{k}-\frac{1}{1}+\frac{1}{9}\)

Alan  Apr 3, 2023
 #8
avatar+1124 
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Thank you Alan....I'll go through this again carefully...much appreciated

juriemagic  Apr 3, 2023
 #3
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The limits for the summations are the same so why not simply say

\(\displaystyle \sum^{10}_{y=3}\frac{1}{y-2}-\sum^{10}_{y=3}\frac{1}{y-1}=\sum^{10}_{y=3}\left\{ \frac{1}{y-2}-\frac{1}{y-1}\right\} \\ =(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+\dots+(1/8-1/9) \\ = 1-1/9=8/9.\)

 Apr 3, 2023
 #5
avatar+1124 
0

Hi guest,

 

yes this was basically my approach, then Alan gave me a more elegant way, which I actually like, but just fail to see how he got certasin values...Thank you for giving your input as well, appreciated..

juriemagic  Apr 3, 2023
 #6
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0

Hardly the same. Neither series has been summed,

The way that it's set out, the halves cancel, the thirds cancel all the way up to the eighths cancelling, leaving just the 1 and the1/9.

Guest Apr 3, 2023
 #7
avatar+1124 
0

Thank you

juriemagic  Apr 3, 2023

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