HELP on these Challenging Questions (1 Viewer)

[Bubbles531]

Hello, I will have an exam soon, can anyone please help me on these questions ?​

​

1.Let Xn=1/(n+1) +1/(n+2) +...+ 1/(2n)

Find the limit of this as n approaches infinity


2.Evaluate the sum of the series

​

∞​

E 1/(k(k+1)(k+2))​

k=1​

​

3. For which z that is an element of C does the series​

∞​

E (z/(z+1))^k converge ?​

=0​

 

[blakwidow]

These are hard. Do you know from what book these questions are from?

 

[airie]

Hello, I will have an exam soon, can anyone please help me on these questions ?​

​

1.Let Xn=1/(n+1) +1/(n+2) +...+ 1/(2n)

Find the limit of this as n approaches infinity


2.Evaluate the sum of the series

​

∞​

E 1/(k(k+1)(k+2))​

k=1​

​

3. For which z that is an element of C does the series​

∞​

E (z/(z+1))^k converge ?​

=0​

1. As n approaches infinity, so do n+1, n+2, ..., 2n, thus 1/(n+1), 1/(n+2), ..., 1/2n all approach zero; therefore the sum Xn approaches zero.

2. Each term 1/k(k+1)(k+2) can be written as 1/k * (1/(k+1) - 1/(k+2), so the sum equals:
ee[/colour]1/1 * (1/2 - 1/3) + 1/2 * (1/3 - 1/4) + ...
= 1/1 * 1/2 - 1/1 * 1/3
eeeeeeeeee+ 1/2 * 1/3 - 1/2 * 1/4
eeeeeeeeeeeeeeeeeeee+ 1/3 * 1/4 - 1/3 * 1/5 ...
= 1/1 * 1/2 - 1/(1*2*3) - 1/(2*3*4) - ...

Add this to the original expression of the sum, 1/(1*2*3) + 1/(2*3*4) +..., one obtains:
2*Sum = 1/1 *1/2 = 1/2,

Therefore the sum equals 1/4.

3. The sum is 1 + z/(z+1) + z²/(z+1)² + z³/(z+1)³ +...
ie. this is a GP with common ratio z/(z+1).

As you know, a GP is convergent iff the common ratio is in the open invertal (-1,1), just solve -1 < z/(z+1) < 1.

 

[bos1234]

doesn't converge mean the limiting sum?

 

[pLuvia]

doesn't converge mean the limiting sum?

A converging series is one that has a limiting sum so yes

So airie is just using the definition of a GP where the r>|1| for it to be convergent

 

[bos1234]

2. Each term 1/k(k+1)(k+2) can be written as 1/k * (1/(k+1) - 1/(k+2), so the 1.

So after you find z you can find limitng sum?

 

[bos1234]

1/1 * (1/2 - 1/3) + 1/2 * (1/3 - 1/4) + ...
= 1/1 * 1/2 - 1/1 * 1/3
eeeeeeeeee+ 1/2 * 1/3 - 1/2 * 1/4
eeeeeeeeeeeeeeeeeeee+ 1/3 * 1/4 - 1/3 * 1/5 ...
= 1/1 * 1/2 - 1/(1*2*3) - 1/(2*3*4) - ...

.

how you go from that step to the next?

 

[AMorris]

"how you go from that step to the next?"

airie stated that
1/(k(k+1)(k+2)) = 1/k(1/(k+1)-1/(k+2)) (easily verified by expanding the RHS)


Now by a similar method:

1/(k(k+1)(k+2)) = 1/(k+2)(1/k - 1/(k+1)) (easily verified)


now she has just grouped the two terms which are in teh same column

so - 1/1 * 1/3 + 1/2 * 1/3 = 1/3(1/2 - 1/1) = -1/3(1/1 - 1/2) = -1/(1*2*3)

where the last equality is derived from the second identity.


"1. As n approaches infinity, so do n+1, n+2, ..., 2n, thus 1/(n+1), 1/(n+2), ..., 1/2n all approach zero; therefore the sum Xn approaches zero."


Unfortunately this is wrong I think.

1/(n+1) > 1/2n
1/(n+2) > 1/2n
...
1/2n = 1/2n

And there are n terms in our sum X_n. So:

X_n > 1/2n + 1/2n + ... + 1/2n
= n(1/2n)
=1/2

and this is irrespective of how large n is. In addition:

1/(n+1) < 1/n
1/(n+2) < 1/n
...
1/2n < 1/n

Adding these n terms together yields:

X_n < 1/n + 1/n + ... + 1/n
= n(1/n)
= 1

so we know that for all n (there may be equality for small n), 1 > X_n > 1/2.

The answer is log2 (natural log) I'm pretty sure but I have no elementary way to derive that.

 

[bos1234]

thanks mr morris

 

[airie]

"1. As n approaches infinity, so do n+1, n+2, ..., 2n, thus 1/(n+1), 1/(n+2), ..., 1/2n all approach zero; therefore the sum Xn approaches zero."


Unfortunately this is wrong I think.

1/(n+1) > 1/2n
1/(n+2) > 1/2n
...
1/2n = 1/2n

And there are n terms in our sum X_n. So:

X_n > 1/2n + 1/2n + ... + 1/2n
= n(1/2n)
=1/2

and this is irrespective of how large n is. In addition:

1/(n+1) < 1/n
1/(n+2) < 1/n
...
1/2n < 1/n

Adding these n terms together yields:

X_n < 1/n + 1/n + ... + 1/n
= n(1/n)
= 1

so we know that for all n (there may be equality for small n), 1 > X_n > 1/2.

The answer is log2 (natural log) I'm pretty sure but I have no elementary way to derive that.

*kicks self for being lazy*

Thanks for pointing that out, Morris