# limit at infinity of sum of arithmetic sequence

• Jan 12th 2007, 06:24 AM
ling_c_0202
limit at infinity of sum of arithmetic sequence
First,
a = (1 + 2 + 3 + ... + n) / (n^2)

When I try to find the limit of a when n tends to infinity,

I have tried two ways:

1. I converted a to
2 - [(1 / 2) ^ (n) ] and get the value 2.

2. I make it to

a = 1 / n^2 + 2 / n^2 + 3 / n^2 + ... + n/ n^2 and then take the limit.
but this give an answer 0.

Why? I know there must be something wrong.
• Jan 12th 2007, 06:34 AM
earboth
Quote:

Originally Posted by ling_c_0202
First,
a = (1 + 2 + 3 + ... + n) / (n^2)

When I try to find the limit of a when n tends to infinity,

I have tried two ways:

1. I converted a to
2 - [(1 / 2) ^ (n) ] and get the value 2.

2. I make it to

a = 1 / n^2 + 2 / n^2 + 3 / n^2 + ... + n/ n^2 and then take the limit.
but this give an answer 0.

Why? I know there must be something wrong.

Hello,

use the property:

1 + 2 + 3 + ... + n = 1/2*n*(n+1)

Using the usual methods you'll get 1/2 as the limit if n approaches infinity.

EB
• Jan 12th 2007, 06:55 AM
ling_c_0202
Quote:

Originally Posted by earboth
Hello,

use the property:

1 + 2 + 3 + ... + n = 1/2*n*(n+1)

Using the usual methods you'll get 1/2 as the limit if n approaches infinity.

EB

But what's wrong with my second method?
• Jan 12th 2007, 06:58 AM
topsquark
Quote:

Originally Posted by ling_c_0202
But what's wrong with my second method?

The sum 1/n^2 + ... n/n^2 is an infinite sum. You can't just use the limit of the last term to find the limit of the sum.

-Dan
• Jan 12th 2007, 07:45 AM
CaptainBlack
$\displaystyle \lim_{n \to \infty} \frac{\sum_{r=1}^n r}{n^2}=\lim_{n \to \infty}\frac{n(n+1)/2}{n^2}$$\displaystyle =\lim_{n \to \infty} \left[n^2/(2n^2)+n/(2n^2)\right]=\lim_{n \to \infty}\left[1/2+1/(2n)\right]=1/2$