Method for finding the general term of a sequence

[tsamocki]

Homework Statement

Find the general term of the sequence, starting with n=1, determine whether the sequence converges, and if so find its limit.

(a) {(1/2), (3/4), (5/6), (7/8), ...}

(b) {(1-(1/2), ((1/3)-(1/2)), ((1/3)-(1/4)), ((1/5)-(1/4)),...}

Homework Equations

My text only provides a few (basic) examples, but they use a term number/term table.

The Attempt at a Solution

I know that the general term for (a)= {(2n-1)/(2n)}n=1 to infinity, and that it converges at its limit of 1, but i was able to figure it out by playing around with basic arithmetic, then finding the limit. For a more complex sequence such as (b), i really don't know where to begin.

Is there a method for finding the general terms of sequences/infinite series?

 

[tiny-tim]

hi tsamocki!

I know that the general term for (a)= {(2n-1)/(2n)}n=1 to infinity, and that it converges at its limit of 1, but i was able to figure it out by playing around with basic arithmetic, then finding the limit.

subtract the general term from 1, and prove that it converges to 0

For a more complex sequence such as (b), i really don't know where to begin.

the odd and even terms seem to be unconnected, so i'd treat them as two different series, and find the limit of each

 

[I like Serena]

Homework Statement

Find the general term of the sequence, starting with n=1, determine whether the sequence converges, and if so find its limit.

(a) {(1/2), (3/4), (5/6), (7/8), ...}

(b) {(1-(1/2), ((1/3)-(1/2)), ((1/3)-(1/4)), ((1/5)-(1/4)),...}

Homework Equations

My text only provides a few (basic) examples, but they use a term number/term table.

The Attempt at a Solution

I know that the general term for (a)= {(2n-1)/(2n)}n=1 to infinity, and that it converges at its limit of 1, but i was able to figure it out by playing around with basic arithmetic, then finding the limit. For a more complex sequence such as (b), i really don't know where to begin.

Is there a method for finding the general terms of sequences/infinite series?

It starts with just looking at the pattern.
In (b) you'll see that each consecutive entry contains the numbers n and (n+1).
What may help is the fact that you can use (-1)^n to be able to toggle the sign of the terms.
Writing out the formula will tell you where it converges.

 

[tsamocki]

hi tsamocki!


subtract the general term from 1, and prove that it converges to 0

The solution i provided for (a) is correct, i was just putting it in brace notation.

the odd and even terms seem to be unconnected, so i'd treat them as two different series, and find the limit of each

I cannot find the general term for (b), but it appears that they are getting closer and closer to zero. In order to find the limit, i would first have to find the general term, correct?

 

[tsamocki]

It starts with just looking at the pattern.
In (b) you'll see that each consecutive entry contains the numbers n and (n+1).
What may help is the fact that you can use (-1)^n to be able to toggle the sign of the terms.
Writing out the formula will tell you where it converges.

The osculating signs are a factor of (-1)^(n+1) because a1 is positive, but i am still having trouble finding the rest of the product for the GT.

Do you really have to just play around with it until you find one that works? If so, how can you prove the GT applies when n= 1,000,000 or something even larger?

 

[Dick]

The osculating signs are a factor of (-1)^(n+1) because a1 is positive, but i am still having trouble finding the rest of the product for the GT.

Do you really have to just play around with it until you find one that works? If so, how can you prove the GT applies when n= 1,000,000 or something even larger?

You can't prove anything. They only gave you four terms. You have to guess. How about something simple like (-1)^(n+1)*(1/n-1/(n+1))?

 

[tsamocki]

You can't prove anything. They only gave you four terms. You have to guess. How about something simple like (-1)^(n+1)*(1/n-1/(n+1))?

Yes, that works. Is it possible to have more than one solution?

For example, your solution gave me an idea: (-1)^(n+1)*((1/n)-(1/n+1)) seems to work as well.

 

[I like Serena]

Yes, that works. Is it possible to have more than one solution?

For example, your solution gave me an idea: (-1)^(n+1)*((1/n)-(1/n+1)) seems to work as well.

Isn't that the same solution?
You only changed the parentheses.
Actually you can't leave out the parentheses in 1/(n+1), because without them the solution is wrong. That is because 1/n+1 means (1/n)+1.

 

[tsamocki]

Isn't that the same solution?
You only changed the parentheses.
Actually you can't leave out the parentheses in 1/(n+1), because without them the solution is wrong. That is because 1/n+1 means (1/n)+1.

Yes, i did mistakenly leave out that parentheses.

It should look like this: (-1)^(n+1)*((1/n)-(1/(n+1))) which is not the same answer as what Dick provided because i am subtracting the (1/(n+1)) from (1/n).

Unless of course i am mistaking something (which is very possible lol).

Am i wrong to assume that the general terms are not unique to a specific sequence?

 

[Dick]

Yes, i did mistakenly leave out that parentheses.

It should look like this: (-1)^(n+1)*((1/n)-(1/(n+1))) which is not the same answer as what Dick provided because i am subtracting the (1/(n+1)) from (1/n).

Unless of course i am mistaking something (which is very possible lol).

Am i wrong to assume that the general terms are not unique to a specific sequence?

I meant to write, and thought I did, exactly what you are writing. Now can you show the limit is zero?

 

[tsamocki]

I meant to write, and thought I did, exactly what you are writing. Now can you show the limit is zero?

Almost; after simplifying , i am still left with a numerator of (-1)^(n+1), which confuses me. Is it safe to assume that as n -> infinity, the +1 is trivial, and can instead take the numerator to be simply (-1)^n and the denominator to be n^2-n (which is easy to solve)?

P.S. The original solution you gave me seemed to give me the same terms, or am i simply adding wrong?

 

[Dick]

Almost; after simplifying , i am still left with a numerator of (-1)^(n+1), which confuses me. Is it safe to assume that as n -> infinity, the +1 is trivial, and can instead take the numerator to be simply (-1)^n and the denominator to be n^2-n (which is easy to solve)?

P.S. The original solution you gave me seemed to give me the same terms, or am i simply adding wrong?

The denominator is n^2+1 isn't it? Why don't you show just what you got when you simplified or whatever you did? You really don't have to worry that much about the (-1)^(n+1) part. If |a_n| converges to zero, then a_n converges to zero. You can show that with a squeeze theorem argument.

 

[tsamocki]

The denominator is n^2+1 isn't it? Why don't you show just what you got when you simplified or whatever you did? You really don't have to worry that much about the (-1)^(n+1) part. If |a_n| converges to zero, then a_n converges to zero. You can show that with a squeeze theorem argument.

Simplifying the general term, i was left with: -(lim (-1)^(n+1)/(n(n+1))), which when n-> infinity, the limit is zero.

Normally when you have a function that has an exponential sum (^(n+1)), wouldn't you make the entire limit be an exponent of e?

I did forget the squeezing theorem :-*

 

[Dick]

Simplifying the general term, i was left with: -(lim (-1)^(n+1)/(n(n+1))), which when n-> infinity, the limit is zero.

Normally when you have a function that has an exponential sum (^(n+1)), wouldn't you make the entire limit be an exponent of e?

I did forget the squeezing theorem :-*

No, you've got it right. The limit is between -1/(n*(n+1)) and +1/(n*(n+1)) by squeeze. I did a typo when is said the denominator is n^2+1. Sorry. I meant n^2+n.