A couple of Infinite Series questions

[Ubinunc]

a couple of "Infinite Series" questions...

Hi,

I'm trying to solve some problems in "Stroud's Engg mathematics"...

I'm stuck with these 2 questions:

Σ (r=0→∞) (2r) / (r+1)!

and

Σ (r=0→n) (2r-1) / r(r+1)(r+2)

the 1st question converges to 2. My 1st try is divide everything the numerator and denominator by r, but I cannot show that (r+1)!/r converges to 1. (or am I missing something?)

for the second question, I was asked to derive a formula for getting the sum for n terms of the series. I'm clueless where to start.

Any help greatly appreciated.

 

[g_edgar]

Can you do $\sum_{r=1}^\infty \frac{1}{(r+1)!}$ ? Can you do $\sum_{r=1}^\infty \frac{r+1}{(r+1)!}$ ? Will these two help with the one you want?

 

[Ubinunc]

Thanks edgar.

My initial approach to the 1st question is right, (r+1)!= (r+1) (r) (r-1)... dividing by r, then it all reduces to 1. It's the 2nd one that I can't figure out.

 

[srijithju]

If the sum is for values of r from 0 to infinity , then the 1st term in your second question is -infinity. So how is the sum supposed to converge ??

 

[srijithju]

Well I just came to know that $\sum_{r=1}^\infty 1/r$ is itself divergent , so I am unsure of your question , it looks like some - $$\infty$$ + $$\infty$$ problem . I am at loss to see what's goin on ... does it mean the sum is undefined (because it doesn't seem to be right to say that the sum is unbounded)?

By the way as regards to the proof that $\sum_{r=1}^\infty 1/r$ is divergent , I saw it written that this could be shown by the integral test - just integrate the term inside the sum as if its a function of r , between the limits r= 1 to $$\infty$$.
How does it work ?
I have some doubts regarding this so called integral test.
Even though it seems to be somewhat logical , but how would this integral test (i.e to show that the integral is bounded) be a necessary and sufficient condition for convergence of a series.
Looks to me like it is a sufficient condition (assuming the function is continuous and doesn't take the valuse $$\infty$$ anywhere within the bounds of the limits) , but I am unsure of how it would be necessary if at all it is.

Now if it is not a necessary condition , then that would mean if the integral is unbounded , then still maybe the sum could be bounded.

 

[robert Ihnot]

If we look at$$\int_1^r\frac{1}{r} =ln(r)$$ This integral, which goes to infinity is less than the upper sum from 1 to r, which shows that it also goes to infinity.

 

[Ubinunc]

THanks srijithju for pointing that out...

it should be:

Σ (r=1→n) (2r-1) / r(r+1)(r+2)

and the formula for getting the sum of n terms for this is n(3n+1)/4(n+1)(n+2)

which I don't know how to get.

Also, my solution in the 1st question is wrong... i was testing for convergence, not getting the closed form... which is 2.

 

[HallsofIvy]

Well, if the "closed form" is 2, the fact that is has a closed form means it converges doesn't it?

If you mean showing that
$$\sum_{r= 0}^\infty \frac{r}{(r+1)!}$$
converges without knowing the sum, use the "ratio test"

$$\frac{r+1}{(r+1+1)!}\frac{(r+1)!}{r+1}= \frac{(r+1)!}{(r+2)!}\frac{r+1}{r}$$
$$= \frac{(r+1)!}{(r+2)(r+1)!}\frac{r+1}{r}= \frac{1}{r+2}\frac{r+1}{r}$$
the first fraction, 1/(r+2), goes to 0 as r goes to infinity while the second, (r+1)/r, goes to 1. The entire expression goes to 0 < 1 so this series converges.

 

[srijithju]

If we look at$$\int_1^r\frac{1}{r} =ln(r)$$ This integral, which goes to infinity is less than the upper sum from 1 to r, which shows that it also goes to infinity.

Thanks Robert for the explanation . Yes I was able to show that

$\sum_{i=1}^r 1/i$ > $$\int_1^r\frac{1}{r} =ln(r)$$

But what bothers me is that I had formed some sort of argument in my mind to convince myself that if the integral is bounded for some function f(x) then the sum shall also be bounded iff :
for some 0 < $$\Delta$$ <= 1/2 The function :$$\frac{1}{f(x)}\int_{x-\Delta}^{x+\Delta}{f(y)}$$ has the limit tending to a finite non -zero value as x tends to $$\infty$$Could someone tell me if the integral test is indeed a sufficient condition for the sum to converge so that I can be sure that there is nothing wrong with my argument. (for the case of 1/r we have shown the integral test is necessary )

PS I would like to write down the complete argument here and how i got to the limit condition , but this latext formatting seems very complicated , and i donno how to use it properly

By the way Robert, you said that "if we look at the integral .. this integral is less than the sum" ... how were you able to say that this integral is less than the sum just by looking at it ? I had to take the derivative and use it to see that diff of sum and derivative is an increasing function

 

[robert Ihnot]

Thanks Robert for the explanation . Yes I was able to show that

$\sum_{i=1}^r 1/i$ > $$\int_1^r\frac{1}{r} =ln(r)$$

But what bothers me is that I had formed some sort of argument in my mind to convince myself that if the integral is bounded for some function f(x) then the sum shall also be bounded iff :
for some 0 < $$\Delta$$ <= 1/2 The function :


$$\frac{1}{f(x)}\int_{x-\Delta}^{x+\Delta}{f(y)}$$ has the limit tending to a finite non -zero value as x tends to $$\infty$$


Could someone tell me if the integral test is indeed a sufficient condition for the sum to converge so that I can be sure that there is nothing wrong with my argument. (for the case of 1/r we have shown the integral test is necessary )

PS I would like to write down the complete argument here and how i got to the limit condition , but this latext formatting seems very complicated , and i donno how to use it properly

By the way Robert, you said that "if we look at the integral .. this integral is less than the sum" ... how were you able to say that this integral is less than the sum just by looking at it ? I had to take the derivative and use it to see that diff of sum and derivative is an increasing function

I am just relying on second quarter calculus to state that "The picture," which is undrawn because of the difficulity, indicates obviously that a block, for example, of unit 1, is greater than a curved area that starts at 1 and decreases continuously to 1/2, a unit away. This may not always be a satisfactory answer.

In other words the curve is monotonically decreasing in area while the solid blocks representing units 1, 1/2, etc stay the same, having a base of one unit. (A monotonic decreasing curve always has a negative derivative.) Undoubtedly, it can be shown that a monotonically decreasing curve from point a to b, of width L, such that F(b)<F(a) has an area S, such that F(a)L>S>F(b)L.

 

[robert Ihnot]

I am just relying on second quarter calculus to state that "The picture," which is undrawn because of the difficulity, indicates obviously that a block, for example, of unit 1, is greater than a curved area that starts at 1 and decreases continuously to 2, a unit away. This may not always be a satisfactory answer.

In other words the curve is monotonically decreasing in area while the solid blocks representing units 1, 1/2, etc stay the same, having a base of one unit. (A monotonic decreasing curve always has a negative derivative.) Undoubtedly, it can be shown that a monotonically decreasing curve from point a to b, of width L, such that F(b)<F(a) has an area S, such that F(a)L>S>F(b)L.

You could employ the Mean Value Theorem, which tells us for the continuous curve, there exists c between a and b such that:

$$f(c)' = \fra{(F(a)-F(b)}{(a-b)}$$. Using a and b as before, we have a-b negative, as is f(c)'. So that under the circumstances this tells us that f(c)'[a-b] = area under the curve. c is an interior point and the curve is monotonically decreasing. So this area is less than F(a)(b-a).

 

[Ubinunc]

Well, if the "closed form" is 2, the fact that is has a closed form means it converges doesn't it?

If you mean showing that
$$\sum_{r= 0}^\infty \frac{r}{(r+1)!}$$
converges without knowing the sum, use the "ratio test"

$$\frac{r+1}{(r+1+1)!}\frac{(r+1)!}{r+1}= \frac{(r+1)!}{(r+2)!}\frac{r+1}{r}$$
$$= \frac{(r+1)!}{(r+2)(r+1)!}\frac{r+1}{r}= \frac{1}{r+2}\frac{r+1}{r}$$
the first fraction, 1/(r+2), goes to 0 as r goes to infinity while the second, (r+1)/r, goes to 1. The entire expression goes to 0 < 1 so this series converges.

Yes, it does converge... but how do you get 2? D Alembert's shows only convergence... what's making me pull my hair off is how to get 2? At first look it seems obvious, I actually ran in Mathematica 7, and yep it is 2. But how?

Another problem

$$\sum_{r= 1}^\infty \frac{1}{2r(2r+1)}$$

it is easy to show that it converges, by comparison test with 1/n^2.

But how do you show to what number it converges? It actually converges to 1-log[2].

Sorry if my questions seem elementary, I'm poor at analysis...:(. I think I'm lacking some skills. I only know how to do sums of powers of natural numbers...

 

[Ubinunc]

Sorry if my original question is misleading... I'm not trying to show that the two series converges, but to what it converges to.

Edit/Added:

And then what I did, instead of getting the number/formula to which it converges, I instead checked for convergence.

 

[srijithju]

I am just relying on second quarter calculus to state that "The picture," which is undrawn because of the difficulity, indicates obviously that a block, for example, of unit 1, is greater than a curved area that starts at 1 and decreases continuously to 1/2, a unit away. This may not always be a satisfactory answer.

Thanks again .. what a fool I was .. Rather than considering the area under the curve from i to i+1 , I was considering the area under the curve from i-1/2 to i + 1/2 , that's why I had to take the difference and check for monotonicity and all ... it was pretty obvious .By the way Ubibic , your sum can be solved as follows = $$\sum_{r= 0}^\infty \frac{2r}{(r+1)!}$$

= 2 * $$\sum_{r= 0}^\infty \frac{r+1}{(r+1)!} - \sum_{r= 0}^\infty \frac{1}{(r+1)!}$$

= 2 * $$\sum_{r= 0}^\infty \frac{1}{(r)!} - [ \sum_{r= 0}^\infty \frac{1}{(r)!} - 1/0! ]$$

= 2 ( as $$\sum_{r= 0}^\infty \frac{1}{(r)!}$$ converges - it is equal to e )Regarding ,
$$\sum_{r= 1}^\infty \frac{1}{2r(2r+1)}$$ ,
I have no idea .. all I can figure is this is the difference of reciprocals of even numbers from2 onwards and reciprocals of odd numbers from 3 onwards

 

[srijithju]

Ok .. I got how to find the sum :

$$\sum_{r= 1}^\infty \frac{1}{2r(2r+1)}$$

$$\sum_{r= 1}^\infty \frac{1}{2r} - \sum_{r=1}^\infty \frac{1}{(2r+1)}$$

Now consider the series expansion of log(1 + x )

and in this expansion substitute x=1 .

You will get ur answer
=

 

[Petek]

THanks srijithju for pointing that out...

it should be:

Σ (r=1→n) (2r-1) / r(r+1)(r+2)

and the formula for getting the sum of n terms for this is n(3n+1)/4(n+1)(n+2)

which I don't know how to get.

Also, my solution in the 1st question is wrong... i was testing for convergence, not getting the closed form... which is 2.

Here's a hint how to show that the above sum equals n(3n+1)/4(n+1)(n+2): Use partial fractions and solve the following expression:

$$\frac{2r - 1}{r(r+1)(r+2)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2}$$

for A, B and C. Then plug this into the sum and simplify the resulting expression to get the desired result.

 

[Ubinunc]

Srijithju

2*(e - (e-1))

Yikes ! This was the hint g_edgar was tryin to show me, and I didn' know

Oh man, I need to familiarize myself with expansions (...log (1+x))

Thanks Pete K for the hint. Halls and Rob