Sequence

speedx

speedx

@speedx-a8Holq • Oct 25, 2024
TEST THE CONVERGENCE OF THE SERIES
1+ 2/1! + 2^2/2! + 2^3/3! +.............

show by steps 😒
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  • Saandeep Sreerambatla

    Saandeep Sreerambatla

    @saandeep-sreerambatla-hWHU1M • 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?
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  • speedx

    speedx

    @speedx-a8Holq • Aug 11, 2009

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

    mech_guy

    @mech-guy-TMwuj2 • 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.
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  • Anil Jain

    Anil Jain

    @CrazyBoy • Aug 12, 2009

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

    -CB
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  • mech_guy

    mech_guy

    @mech-guy-TMwuj2 • 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
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