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Limits to e!

Okay, I understand that the lim(n->inf) (1+(1/n))^n = e
but I was just wondering what these limits where..

lim(n->inf) (2+(1/n))^n = ??

lim(n->inf) (1+(5/n))^n = ?? e^5n ??


I guess it's just trivial, but any help is great! :smile:

Thanks
First one would, for obvious reasons, be infinite, as (2+1/n)^n > 2^n. Second would be e^5, which can be proved by a similar method to the fact that e = lim (1 + 1/n)^n. :smile:
sleeping in
Okay, I understand that the lim(n->inf) (1+(1/n))^n = e
but I was just wondering what these limits where..

lim(n->inf) (2+(1/n))^n = ??

lim(n->inf) (1+(5/n))^n = ?? e^5n ??


I guess it's just trivial, but any help is great! :smile:

Thanks

really? or do you just believe it
Reply 3
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sleeping in
OP
So what about ((3/5)+(1/n))^n as n->inf

thanks.
sleeping in
So what about ((3/5)+(1/n))^n as n->inf

thanks.

What do you think?

(What happens when n > 5?)
Reply 5
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sleeping in
OP
I'm guessing it will tend to zero then?
sleeping in
I'm guessing it will tend to zero then?

Correct. Why?
Reply 7
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sleeping in
OP
Well for n<5, we have terms greater than 1, however when we start looking at n>5 our 1/n gets progressively smaller and as we don't have '1+' are expression will get smaller and smaller, as soon (3/5 + 1/n) << 1.

Now, if we are looking at limits of sequences, can we use the same argument? considering for example,

(a)_n>1 = (3/5 + (1/n))^n

_ denotes subscript,

Thanks
sleeping in
Well for n<5, we have terms greater than 1, however when we start looking at n>5 our 1/n gets progressively smaller and as we don't have '1+' are expression will get smaller and smaller, as soon (3/5 + 1/n) << 1.

Well, basically. For n > 5, we have 1/n < 1/5 and so 0 < (3/5 + 1/n)^n < (4/5)^n which converges to 0 (geometric progression). Thus (3/5 + 1/n)^n converges (comparison test) to 0 (squeeze / sandwich theorem / other silly names).

sleeping in
Now, if we are looking at limits of sequences, can we use the same argument? considering for example,

(a)_n>1 = (3/5 + (1/n))^n

_ denotes subscript,

Thanks

You mean you want to find lim⁡n→∞an\displaystyle\lim_{n\to\infty} a_n where an=(35+1n)na_n = (\frac{3}{5} + \frac{1}{n})^n ? Yes, same argument - the limit's 0.
Reply 9
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OP
generalebriety
Well, basically. For n > 5, we have 1/n < 1/5 and so 0 < (3/5 + 1/n)^n < (4/5)^n which converges to 0 (geometric progression). Thus (3/5 + 1/n)^n converges (comparison test) to 0 (squeeze / sandwich theorem / other silly names).


You mean you want to find lim⁡n→∞an\displaystyle\lim_{n\to\infty} a_n where an=(35+1n)na_n = (\frac{3}{5} + \frac{1}{n})^n ? Yes, same argument - the limit's 0.


That's great, thanks a lot! :biggrin:
No problem. :smile:

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