+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.
+1124 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..
+33661 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).
\(\sum_{k=2}^{k=9}\frac{1}{k}=\sum_{k=1}^{k=9}\frac{1}{k}-\frac{1}{1} =\sum_{k=1}^{k=8}\frac{1}{k}-\frac{1}{1}+\frac{1}{9}\)
+1124 Thank you Alan....I'll go through this again carefully...much appreciated
The limits for the summations are the same so why not simply say
\(\displaystyle \sum^{10}_{y=3}\frac{1}{y-2}-\sum^{10}_{y=3}\frac{1}{y-1}=\sum^{10}_{y=3}\left\{ \frac{1}{y-2}-\frac{1}{y-1}\right\} \\ =(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+\dots+(1/8-1/9) \\ = 1-1/9=8/9.\)
+1124 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..
