1. ## sequences

Could someone just tell me if the following sequences converge or not:

Let an = the summation of (1/x) beginning with x = n+1 and ending with 2n.
For example, if n =2, x =3, then the sequence is (1/3) + (1/4).

Let bn = the summation of (1/x) beginning with x = n+1 and ending with pn where p is a positive integer greater than one.
For example, if n =2, x =3, and p =3, then the sequence is (1/3) + (1/4) + (1/5) + (1/6).

It seems to me that both sequences are increasing and bounded above, which would make them convergent. Is this true or not?

2. Originally Posted by PvtBillPilgrim
Could someone just tell me if the following sequences converge or not:

Let an = the summation of (1/x) beginning with x = n+1 and ending with 2n.
For example, if n =2, x =3, then the sequence is (1/3) + (1/4).

Let bn = the summation of (1/x) beginning with x = n+1 and ending with pn where p is a positive integer greater than one.
For example, if n =2, x =3, and p =3, then the sequence is (1/3) + (1/4) + (1/5) + (1/6).

It seems to me that both sequences are increasing and bounded above, which would make them convergent. Is this true or not?
The first is as it is of order:

$\int_{n}^{2n}1/x ~dx = \ln(2)$,

I will leave it to you to sort out the details of the limit.

The second I can't tell what the upper limit of summation is supposed to be.

RonL

3. I'm not sure if you understand what I'm saying.

I have two sequences:

the first one is the summation symbol (sigma) with 2n on top and i = (n+1) on the bottom (I changed x to i to eliminate possible confusion), i am summing (1/i) with those conditions

the second one is the summation symbol (sigma) with pn on top and i = (n+1) on the bottom where p is a positive integer, i am summing (1/i) with those conditions

are these convergent?

4. Have you noticed that:
$\begin{array}{l}
a_5 = \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{{10}} \\
\frac{1}{2} = \frac{5}{{10}} < \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{{10}} < \frac{5}{5} = 1 \\
\end{array}$
?

So we have:
$\frac{1}{2} < a_n = \sum\limits_{k = n + 1}^{2n} {\frac{1}{k}} < 1$.

5. Originally Posted by PvtBillPilgrim
I'm not sure if you understand what I'm saying.

I have two sequences:

the first one is the summation symbol (sigma) with 2n on top and i = (n+1) on the bottom (I changed x to i to eliminate possible confusion), i am summing (1/i) with those conditions
That is what I was addressing.

RonL

6. Originally Posted by PvtBillPilgrim
the second one is the summation symbol (sigma) with pn on top and i = (n+1) on the bottom where p is a positive integer, i am summing (1/i) with those conditions

are these convergent?
$\sum_{r=n+1}^{p\times n} 1/r,\ p \in \mathbb{N},\ p \ge 2$.

Yes is convergent to $\ln(p)$ for the same reason that the first
one converges to $\ln(2)$

RonL

7. I have a question. Am I the only one here that has misgivings about the captain’s response?
I have no doubt that that the value of the limit he gave is correct.
But it seems to me that the question was about showing that the sequence has a limit.
If I had set this question for a class, I must confess that I think that I would not accept that as an answer.
I would truly like some input on this question.

8. Originally Posted by PvtBillPilgrim
Could someone just tell me if the following sequences converge or not:

Let an = the summation of (1/x) beginning with x = n+1 and ending with 2n.
For example, if n =2, x =3, then the sequence is (1/3) + (1/4).
$\frac{1}{n+1}+...+\frac{1}{2n} = \frac{1}{n}\left( \frac{1}{1+\frac{1}{n}} + ... + \frac{1}{1+\frac{n}{n}} \right) = \frac{1}{n} \sum_{k=1}^n \frac{1}{1+\frac{k}{n}}$
This the the Riemann sum with equal sub-intervals so:
$\lim \ \frac{1}{n} \sum_{k=1}^n \frac{1}{1+\frac{k}{n}} = \int_0^1 \frac{dx}{x+1} = \ln 2$

9. Originally Posted by Plato
I have a question. Am I the only one here that has misgivings about the captain’s response?
I have no doubt that that the value of the limit he gave is correct.
But it seems to me that the question was about showing that the sequence has a limit.
If I had set this question for a class, I must confess that I think that I would not accept that as an answer.
I would truly like some input on this question.
If you read what I wrote I was leaving it to the poster to work out the detail.

The idea is that:

$
\sum\limits_{k = n + 1}^{2n} {\frac{1}{k}} = \int_n^{2n} 1/x ~dx + O(1/n)
$

With some effort you can establish explicit bounds on the error, which is
what I want the poster to do if he cannot see that it is $O(1/n)$ which
is all we need to establish what the limit actually is.

Think of this as an illustration of how we really do these things.

RonL