Looks like the Harmonic series

[Punkyc7]

does let z_{n}=

\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n}

does z_{n} converge or diverge..


I want to say it diverges because it looks like the Harmonic series

 

[Punkyc7]

yep I am positive

 

[Mark44]

does let z_{n}=

\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n}

does z_{n} converge or diverge..


I want to say it diverges because it looks like the Harmonic series

Each term in the sum above is >= 1/(2n), which is the smallest term, and there are n terms. This means that z_n >= n/(2n) = 1/2.

Also, each term in the sum is <= 1/(n + 1), which is the largest term, and there are still n terms. This means that z_n <= n/(n + 1).

One of the tests you have learned is applicable here.

 

[Punkyc7]

does that mean it converges?

 

[Mark44]

Write down a few terms in your sequence.
z₁ = 1/2
z₂ = 1/3 + 1/4
z₃ = 1/4 + 1/5 + 1/6

What can you say about
\lim_{n \to \infty} z_n~?

Note that I added some to my previous response.

 

[Dickfore]

<br /> \begin{array}{rcl}<br /> z_{n + 1} - z_{n} &amp; = &amp; \left( \frac{1}{n + 2} + \ldots + \frac{1}{2 n} + \frac{1}{2 n + 1} + \frac{1}{2 n + 2} \right) \\<br /> <br /> &amp; - &amp; \left(\frac{1}{n + 1} + \frac{1}{n + 2} + \ldots + \frac{1}{2 n} \right) \\<br /> <br /> &amp; = &amp; \frac{1}{2 n + 1} + \frac{1}{2 n + 2} - \frac{1}{n + 1} \\<br /> <br /> &amp; = &amp; \frac{1}{2 n + 1}- \frac{1}{2 n + 2} &gt; 0<br /> \end{array}<br />

Post #4 holds the other clue.

 

[Punkyc7]

as n goes to infinity isn't the limit 1? because of zn <= n/(n + 1).

 

[Mark44]

About all you can say is that 1/2 <= z_n <= 1, but that doesn't mean that the limit has to be either of the endpoints.

 

[Punkyc7]

since its bounded and increasing it converges doesnt

 

[Dickfore]

as n goes to infinity isn't the limit 1? because of zn <= n/(n + 1).

no. look at, for example:
<br /> z_{2 n} = \left( \frac{1}{2n + 1} + \ldots + \frac{1}{3 n} \right) + \left(\frac{1}{3 n + 1} + \ldots + \frac{1}{4 n}\right) \le \frac{n}{2 n + 1} + \frac{n}{3 n + 1} = \frac{n (5 n + 2)}{(2 n + 1)(3 n + 1)}<br />
<br /> z_{2 n} \le \frac{5 n^2 + 2 n}{6 n^2 + 5 n + 1} = \frac{1}{\frac{6 n^2 + 5 n + 1}{5 n^2 + 2 n}} = \frac{1}{\frac{6}{5} + \frac{\frac{13 n}{5} + 1}{n (5 n + 2)}} \le \frac{1}{\frac{6}{5}} = \frac{5}{6}<br />

Thus, there is a subsequence that is definitely not within a neighborhood \epsilon = 1/5 of 1. So, 1 is not a limit of the sequence.

since its bounded and increasing it converges doesnt

yes, but it's actually decreasing.

 

[Punkyc7]

ok since its monotonic decreasing and bounded it is convergent

 

[Dick]

yes, but it's actually decreasing.

It is NOT decreasing, it's increasing. I'd suggest to Punkyc7 not to believe everything everyone tells you.

 

[Dickfore]

ok since its monotonic decreasing and bounded it is convergent

yes, and the limit is somewhere between 1/2 and 5/6. Actually, the exact sum is expressible via elementary functions. Do you know what it is?

 

[Dickfore]

It is NOT decreasing, it's increasing. I'd suggest to Punkyc7 not to believe everything everyone tells you.

read post #6.

 

[Dick]

read post #6.

I did. And I have no idea what you are doing. z_1=1/2, z_2=7/12, z_3=37/60. Now you read post #6 and tell me what's wrong with it.

 

[Dickfore]

there's an extra term -1/n that does not belong there.

 

[Dick]

there's an extra term -1/n that does not belong there.

Perhaps. I didn't read it THAT carefully. But I do know z_n is an increasingly accurate estimate to log(2). And all of the z_n are underestimates. Think of it as like an integral estimate.

 

[Dickfore]

Perhaps. I didn't read it THAT carefully.

I corrected it.

But I do know z_n is an increasingly accurate estimate to log(2). And all of the z_n are underestimates. hink
of it as like an integral test.

Look at my post #13.

 

[Dick]

I corrected it.



Look at my post #13.

Ok, so you know what it is. That's good. I'd still suggest to Punkyc7 to think more about the problem. And maybe for you to maybe provide less details until Punkyc7 does that.

 

[HallsofIvy]

The original sum is a sum of positive terms. If we are always adding a positive number, the sum must be increasing.

 

[Dickfore]

The original sum is a sum of positive terms. If we are always adding a positive number, the sum must be increasing.

But, we are not adding a positive number in addition to the same positive numbers from the previous term. Thus, your logic is flawed.