# How to find limit of arithmetic series?

• kenvin100
AND the denominator going to infinity. That's the case here."0" is the result of the numerator approaching 0 and the denominator approaching infinity. That's not the case here.In summary, the problem is to find the limit of the series (1 + 3 + 5 + 7 + ... (2n-1)) divided by the expression (n+1) - [(2n+1)/2]. The series is not an arithmetic series and does not converge. To solve the limit, it is helpful to divide each term by the highest power of n in the denominator, which results in an easier expression to work with. In this case, the limit approaches infinity as both
kenvin100
Find the limit of the following series:

lim (n-->infinity) 1 + 3 + 5 + 7 + ... (2n-1) / (n+1) - [(2n+1)/2]

3. From what I kno this is an arithmetic series, meaning I must use that arithmetic series formula. so its (first term + last term / 2 times the number of terms) n^2..

now my teacher did another step, and this is where I am lost:

n^2/n+1 - (2n+1)/n

I can solve it from here but can someone be kind enough to explain the above part to me. (by the way, answer is -3/2)

I also have another question, do i need to divide by highest power of n (in this case) of denominator to all numbers when n approaches infinity? So i need to do that everytime?

Any comments are most appreciated and I thank you for your time :-)

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Can you post your question again, with more brackets :( Its hard to tell what you mean.

In any case, $$1+3+5...+ n = \left( \frac{n+1}{2}\right)^2$$.

And when you have the quotient of polynomials, to find the limit always divide through by the highest power of x.

yes okay sorry

[1 + 3 + 5 ...(2n-1)] / (n+1) - [(2n+1)/2]

so 1 is the first term, and 2n-1 is the last term of the series. I don't know what the other part after the subtraction sign is...

i hope this helped

kenvin100 said:
Find the limit of the following series:

lim (n-->infinity) 1 + 3 + 5 + 7 + ... (2n-1) / (n+1) - [(2n+1)/2]

3. From what I kno this is an arithmetic series, meaning I must use that arithmetic series formula. so its (first term + last term / 2 times the number of terms) n^2..

now my teacher did another step, and this is where I am lost:

n^2/n+1 - (2n+1)/n

I can solve it from here but can someone be kind enough to explain the above part to me. (by the way, answer is -3/2)
No, this is not an arithmetic series! An arithmetic series, where the numbers are getting larger and larger, can't possibly converge. Here you might have an arithmetic series divided by something. It is not clear whether you mean
$$\frac{1+ 3+ 5+ 7+ \cdot\cdot\cdot+ (2n-1)}{(n+1) - \frac{2n+1}{2}}$$
or
$$\frac{1+ 3+ 5+ 7+ \cdot\cdot\cdot+ (2n-1)}{(n+1)}- \frac{2n+1}{2}$$

Unfortunately, I can't see that either of those converges- and, in fact, what your teacher gives does not converge!

I also have another question, do i need to divide by highest power of n (in this case) of denominator to all numbers when n approaches infinity? So i need to do that everytime?

Any comments are most appreciated and I thank you for your time :-)
Well, you don't NEED to- if you can find some other way to do the limit. But it certainly is usually the easiest way to do such a limit. The obvious point is that it is easier to work with "0" than with infinity! Dividing each term by the highest power puts n into the denominator of each individual fraction- as n goes to infinity, each fraction goes to 0.

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ye the whole series is divided by n+1..

edit: nvm i solved it...

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the above poster didn't say it, so i will because it just helped me while doing my calculus homework here:

thanks

lim (n→∞) 1 + 3 + 5 + 7 + ... (2n-1) / (n+1) - [(2n+1)/2]

first we find arithmetic sum: Sn=n/2*[2a1+(n-1)*d], where d=2, a1=1

lim (n→∞) n/2*[2+(n-1)*2] / (n+1) - [(2n+1)/2]

lim (n→∞) n/2*[2+2n-2] / (n+1) - [(2n+1)/2]

lim (n→∞) n(n) / (2n-1) / (n+1) - [(2n+1)/2]

lim (n→∞) n^2/ 2(n+1)-2n+1/2

lim(n→∞)n^2/ 2n+2-2n+1/2

lim(n→∞)n^2/ 2/3

2/0

"infinity" is the result

## 1. How do I find the sum of an arithmetic series?

To find the sum of an arithmetic series, you can use the formula: S = (n/2)(a + l), where n is the number of terms, a is the first term, and l is the last term of the series.

## 2. What is the formula for finding the nth term of an arithmetic series?

The formula for finding the nth term of an arithmetic series is: an = a1 + (n-1)d, where a1 is the first term and d is the common difference between terms.

## 3. How do I determine if an arithmetic series is finite or infinite?

An arithmetic series is finite if it has a definite number of terms, and infinite if it continues indefinitely. To determine this, you can look at the given terms and see if there is a pattern or if the series continues indefinitely.

## 4. What is the difference between an arithmetic series and a geometric series?

An arithmetic series has a constant difference between terms, while a geometric series has a constant ratio between terms. In other words, an arithmetic series increases or decreases by a fixed amount each term, while a geometric series increases or decreases by a fixed factor each term.

## 5. Can the limit of an arithmetic series be negative?

Yes, the limit of an arithmetic series can be negative. This depends on the starting term and the common difference between terms. The limit can be positive, negative, or even zero, depending on the values involved.

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