Prove that the sequence converges.

For each positive integer $n$, let $$y_n = 1 + \frac12 + \frac13 + \cdots + \frac1n - \int_1^n \frac{dx}x.$$

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    $\begingroup$ Did you attempt the question? Can you show us your work and where you're stuck? HINT: One approach for this problem would be to prove that the sequence is monotone (decreasing) and bounded. $\endgroup$ – Srivatsan Dec 9 '11 at 3:00
  • $\begingroup$ Very related... $\endgroup$ – J. M. is a poor mathematician Dec 9 '11 at 3:10
  • $\begingroup$ Show the sequence is monotone decreasing. Since the sequence in monotone decreasing, it is Riemann integrable. Since the sequence is Riemann integrable, it is bounded. Since the sequence is bounded and monotone, it is convergent. $\endgroup$ – GreenBeans Dec 9 '11 at 4:42
  • $\begingroup$ This is also related. $\endgroup$ – J. M. is a poor mathematician Dec 9 '11 at 6:41

enter image description here

The blue-curve takes the value $\frac1{k}$ over an interval $[k,k+1)$

The red-curve is given by $f(x) = \frac1{x}$ where $x \in [1,\infty)$

The green-curve takes the value $\frac1{k+1}$ over an interval $[k,k+1)$

The area under the blue-curve represents the sum $\displaystyle \sum_{k=1}^{n} \frac1{k}$

The area under the red-curve is given by the integral $\displaystyle \int_{1}^{n+1} \frac{dx}{x}$

The area under the green-curve represents the sum $\displaystyle \sum_{k=1}^{n} \frac1{k+1}$

Can you now formalize this argument?

  • $\begingroup$ Great. What did you use to create the plot? $\endgroup$ – davidlowryduda Dec 9 '11 at 4:43
  • $\begingroup$ @mixedmath: I typeset (or) draw some nice results and illustrative figures/plots on my macbook and have it preserved. I think this was done on grapher sometime back. $\endgroup$ – user17762 Dec 9 '11 at 4:50
  • $\begingroup$ I see. Thanks. Just good old grapher. $\endgroup$ – davidlowryduda Dec 9 '11 at 6:15

I don't see quite this approach at this or the older question. Along with the sequence $y_n$ as defined, also define a second sequence $$ z_n = y_n - \frac{1}{n}.$$ Note $y_1 = 1,$ while $z_1 = 0.$ So we always have $y_n > z_n.$ Prove that the $y_n$ are decreasing in $n,$ while the $z_n$ are increasing in $n.$ So, all the $y_i$ are greater than all the $z_j.$ But $y_n - z_n = \frac{1}{n}$ becomes arbitrarily small.

I did a little program on a calculator, I get $y_{22} < 0.6,$ while $z_7 > 0.5.$

As I looked at this again, the only calculus part is the necessary proof that $$ \frac{1}{n+1} < \int_n^{n+1} \frac{dx}{x} < \frac{1}{n} $$ which is easy enough.


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