Sequence

speedx · 2009-08-10T14:59:25+00:00

TEST THE CONVERGENCE OF THE SERIES 1+ 2/1! + 2^2/2! + 2^3/3! +............. show by steps 😒

Sequence

speedx

Member

Updated: Oct 25, 2024

Views: 1.3K

TEST THE CONVERGENCE OF THE SERIES
1+ 2/1! + 2^2/2! + 2^3/3! +.............

show by steps 😒

0

Replies

Howdy guest!

Dear guest, you must be logged-in to participate on CrazyEngineers. We would love to have you as a member of our community. Consider creating an account or login.

Replies

Saandeep Sreerambatla

Member • Aug 10, 2009

1+2 + 2+ 1+ 0.333 +0.0833 + 0.01677 + 0.0027

May be solving for few more steps we get a value almost equal to zero as the next term.

Is this the answer expecting?

Are you sure? This action cannot be undone.
Cancel
speedx

Member • Aug 11, 2009

thnks ya .........but if u can SHOW by COMPARISON TEST (infinite seriies ) method it wil be better 😁

Are you sure? This action cannot be undone.
Cancel
mech_guy

Member • Aug 12, 2009
Hi speedx,

speedx
TEST THE CONVERGENCE OF THE SERIES
1+ 2/1! + 2^2/2! + 2^3/3! +.............

show by steps 😒
Series is: A(n) = 2^n / (n!)

To prove its convergence, RATIO TEST will be best which states:
```
If SIGMA(A(n)) be a series with positive terms and
 lim(n->inf) A(n+1)/A(n) = L
Then,
If L<1, then SIGMA(A(n)) converges
If L>1, then SIGMA(A(n)) diverges
If L=1, then SIGMA(A(n)) may converge or diverge.
```
So in this case A(n+1) = 2^(n+1)/(n+1)!
A(n) = 2^n/n!
A(n+1)/A(n) = 2/n -> 0 for lim(n->inf)

Therefore, given series converges for all n.

but if u can SHOW by COMPARISON TEST (infinite seriies ) method it wil be better
Well Comparison Test too can be used here but i think RATIO TEST will be more convenient in this case:-

COMPARISON TEST says:
```
Consider two series A(n) and B(n):-
If for very large values of "n", 
If B(n) >= A(n) then If B(n) converges, A(n) too will converge
If A(n) <= B(n) then If A(n) diverges, B(n) too will diverge.

That is, if larger of the two series "converges" so will the smaller series.
If smaller series "diverges" so will the larger series.
```
Now considering the present series at hand, A(n) = 2^n / n!

We know that for n = 4, "n!" will overtake "2^n", so we can make another series,
B(n) = 2^4 / n
We have made this B(n) so that B(n) >= A(n) for very large values of "n".

Now we know by LIMIT Rule (n-> inf) that 2^4/n converges.
Therefore A(n) also converges.
Are you sure? This action cannot be undone.
Cancel
Anil Jain

Member • Aug 12, 2009

mech_guy... you are a mathematics genius. Hats off to you.

-CB

Are you sure? This action cannot be undone.
Cancel
mech_guy

Member • Aug 13, 2009

Hi CB,

crazyboy
mech_guy... you are a mathematics genius. Hats off to you.

-CB
Oops, that was a bit too much 😀

We all studied this interesting topic of "Convergence of Series" in one of our sem courses. I just googled to revisit what we studied and everything was simple then.

Thanks to "speedx" for posting such question, it made me "keep in touch" with basics 😀

Cheers

Are you sure? This action cannot be undone.
Cancel