Question

Could someone please help me with this question? http://i.imgur.com/Mm9SPFB.jpg

Answer 1

\[\sum_{10}^{1000} \frac{1}{n^2}< \sum_{10}^{1000} \frac{1}{n(n-1)}\]

Answer 2

*

Answer 3

Can you please explain this in detail as I am not very familiar with the sum sign.

Answer 4

*correction in the right hand side \[ \sum_{10}^{1000} \frac{1}{n-1} - \frac{1}{n} = \frac{1}{10} -\frac{1}{9}+\frac{1}{11}-\frac{1}{10}+\frac{1}{12} -\frac{1}{11} +.............+\frac{1}{999}-\frac{1}{998}+\frac{1}{1000}-\frac{1}{999} = \frac{1}{1000}-\frac{1}{9}\]

Answer 5

the sum sign means that starting from the value under the sign we go to the value above it separating each terms by a +

Answer 6

for example \[ \sum^{5}_{1} x = 1+2+3+4+5\]

Answer 7

ok thanks, what happens to the squares as the denominator?

Answer 8

so \[\sum^{1000}_{10} \frac{1}{n^2} = \frac{1}{10^2}+\frac{1}{11^2}+\frac{1}{12^2}....+\frac{1}{1000^2}\]

Answer 9

the inequality on top \[\sum^{1000}_{10} \frac{1}{n^2} <\sum^{1000}_{10} \frac{1}{n (n-1)} \] it can be derived using a host of different methods but the easiest is inspection

Answer 10

\[ \frac{1}{2^2} <\frac{1}{2.1}\]\[ \frac{1}{3^2} <\frac{1}{3.2}\]\[ \frac{1}{4^2} <\frac{1}{4.3}\] here the . implies mulitplication

Answer 11

if i form all such relations and add all the left hand sides together i would get \[ \sum\frac{1}{n^2} <\sum \frac{1}{n(n-1)}\]

Answer 12

Integrate a curve that is always greater than or equal to the area under rectangles described by 1/10^2 + 1/11^2.

Answer 13

So, one idea is to integrate (10+x)^-2 dx from x = 0 to x = 1000, but this is always under or equal to the top of the rectangles.

Answer 14

However, the curve, (9 + x)^-2 upper bounds it as desired.

Answer 15

So, integrate (9+x)^-2 dx from x = 0 to x = 1000.

Answer 16

In which case, you get 1/9 - 1/1009, which is .111 - some small number.

Answer 17

Thus proving it.

Answer 18

since in your question you have been asked to form the sum (the left hand side) from n = 10 to n = 1000, we need \[\sum^{1000}_{10}\frac{1}{n^2}\]
and we have just established that \[\sum^{1000}_{10}\frac{1}{n^2}<\sum^{1000}_{10}\frac{1}{n(n-1)}\]

Answer 19

what we do next is called partial fractions. It is basically breaking one fraction into two or more parts.
\[\frac{1}{n(n-1)}= \frac{1}{n-1} - \frac{1}{n}\]so the right hand side of our inequality becomes \[\sum^{1000}_{10} (\frac{1}{n-1} - \frac{1}{n})\]

Answer 20

\[\sum^{1000}_{10} \frac{1}{n^2}<\sum^{1000}_{10}( \frac{1}{n-1} - \frac{1}{n})\]

Answer 21

look above to the box where i made the correction (the right hand side of that equation is also messed up, below is the final answer), I have opened the right hand side of the equation writing out each individual term. At the end we are left with \[\sum^{1000}_{10} \frac{1}{n-1} - \frac{1}{n} =\frac{1}{9} - \frac{1}{10}+\frac{1}{10} - \frac{1}{11}+\frac{1}{11} - \frac{1}{12}........+\frac{1}{999} - \frac{1}{1000} = \frac{1}{9} - \frac{1}{1000}\] almost all the terms cancel out. using our inequality and and the given sum we can now say that \[\sum^{1000}_{10} \frac{1}{n^2} < \frac{1}{9} - \frac{1}{1000}\]\[\sum^{1000}_{10} \frac{1}{n^2} < 0.11\] thus proved

Answer 22

ok, thanks very much for your help @akitav

Answer 23

It was a pleasure :)

Answer 24

Brilliant!!!