Show that if for , then

Here's what I did:

Then

And since goes to and is a constant greater than 0,

the also goes to

However, I'm pretty sure I need to use the definition of convergence to proves this, but I'm not sure how. Thank you.

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- Feb 16th 2011, 10:42 PMmremwoProving limit
Show that if for , then

Here's what I did:

Then

And since goes to and is a constant greater than 0,

the also goes to

However, I'm pretty sure I need to use the definition of convergence to proves this, but I'm not sure how. Thank you. - Feb 16th 2011, 10:52 PMroninpro
What you have might serve as some intuition, but it does not work as a proof. You used the limit laws to rearrange the expressions, but they only work when the corresponding limits exist. Here is a slightly more formal argument:

Since , we know that given any , when is sufficiently large, we have . Multiplying through by gives . Breaking the absolute values and rearranging, .

This shows that as long as we pick and take , we have , so . - Feb 17th 2011, 12:01 AMmremwo
Great, thanks! I had this all the way down to |a_n - nL| < ne but I always seem to forget that I am able to pick e to fit what I need so long as it is greater than 0.