Sequences/Series, im terrible at them...

[khauna]

anyway, my last question i promise :roll:

we got a mini project thrown on us about sequences but i really dont understand it too well yet, im working on this project and learning it but i need some help and i need to do it soon, our class is almost finished...and im slightly stressed about the final...
This question is long so i could use some help along the way...

Background info:

As n increases, there is very little change in the difference between the sum
1 + 1/2 + 1/3 +...+1/n and the integral \(\displaystyle \int_{1}^{n}(1/x)dx\)

Question is:

by taking f(x) = (1/x) in the double inequality that is used in the proof of the integral test, show that

ln(x+1) < 1+1/2 +1/3+...+1/n - ln n < 1

Thus, the sequence a = 1+1/2 + 1/3 +...+1/n - ln n
is bounded from below and above.

Then we must go on to show that

1/n+1 < \(\displaystyle \int_{n}^{n+1}(1/x)dx\) = ln (n+1) - ln n

and then show that the sequence {a} in the first part of our question is decreasing


- know that a decreasing sequence bounded from below converges...or i think so...

Yea...i know this is a lot...well a lot for me. Any help, pointers, tips, anything will be greatly appreciated, just tell me if i need to clear some things up, i know that could be slightly confusing...

 

[pka]

There is a well known inequality: \(\displaystyle 0 < a < b\Rightarrow \quad \frac{1}{b} \le \frac{{\ln (b) - \ln (a)}}{{b - a}} \le \frac{1}{a}\).
Form which we get:
\(\displaystyle \frac{1}{{k + 1}} \le \ln (k + 1) - \ln (k) \le \frac{1}{k} \Rightarrow \quad \sum\limits_{k = 1}^{n - 1} {\frac{1}{{k + 1}}} \le \sum\limits_{k = 1}^{n - 1} {\left[ {\ln (k + 1) - \ln (k)} \right]} \le \sum\limits_{k = 1}^{n - 1} {\frac{1}{k}}\)
\(\displaystyle \begin{array}{l} \sum\limits_{k = 1}^{n - 1} {\left[ {\ln (k + 1) - \ln (k)} \right]} = \ln \left( n \right) \\ \sum\limits_{k = 1}^{n - 1} {\frac{1}{{k + 1}}} \le \ln (n) \le \sum\limits_{k = 1}^{n - 1} {\frac{1}{k}} \\ \end{array}\)
\(\displaystyle 1 + \sum\limits_{k = 1}^{n - 1} {\frac{1}{{k + 1}}} \le 1 + \ln (n)\,\& \,\ln (n + 1) \le \sum\limits_{k = 1}^n {\frac{1}{k}}\)
Note the change in the last sum from n-1 to n.

 

[khauna]

Okay, I more or less follow your math but im still not certain what you are trying to show me. :?

 

[Deleted member 4993]

Okay, I more or less follow your math but im still not certain what you are trying to show me. :?

He has outlined the complete proof for you.

Take pencil and paper and write it down (don't just stare at the screen) then think.....

 

[khauna]

okay ill do that, ill give it a little time too im still just learning sequences and series