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Real Analysis Proof

  1. Mar 12, 2008 #1
    1. Let r(n) = (1+1/n)^n and t(n) = (1+1/n)^n+1. (Use r(n) converge to e).
    Show that t(n) > r(n) for all n and that lim n->inf(t(n) - r(n)) = 0.
    Show that {tn} is a decreasing sequence with limit e. {Hint: express {(1+1/n-1)/(1+1/n)}^n as (1+a)^n and apply Bernoulli's inequality). Use n=10 to calculate upper and lower estimates for e. How large should n be to estimate e to 3 decimal places?
     
  2. jcsd
  3. Mar 12, 2008 #2

    well about showing that t(n)>r(n) does not appear to be that difficult

    [tex] (1+\frac{1}{n})^{n+1}=(1+\frac{1}{n})^{n}(1+\frac{1}{n})[/tex]
    now since, [tex](1+\frac{1}{n})>1[/tex] we have that

    [tex]t(n)= (1+\frac{1}{n})^{n+1}=(1+\frac{1}{n})^{n})(1+\frac{1}{n})>(1+\frac{1}{n})^{n}=r(n)[/tex]

    now:

    [tex]\lim_{n\rightarrow\infty}(t(n)-r(n))=\lim_{n\rightarrow\infty}((1+\frac{1}{n})^{n}(1+\frac{1}{n}))-\lim_{x\rightarrow\infty}(1+\frac{1}{n})^{n}=\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n}*\lim_{n\rightarrow\infty}(1+\frac{1}{n})-e=e*1-e=0[/tex]

    Now we want to show that [tex]t_n-_1>t_n[/tex] this part is quite easy to show as well

    [tex]\frac{t_n-_1}{t_n}=\frac{(1+\frac{1}{n-1})^{n}}{(1+\frac{1}{n})^{n+1}}=.....=(\frac{n^{2}}{n^{2}-1})^{n}*\frac{n}{n+1}=(\frac{n^2-1+1}{n^{2}-1})^{n}\frac{n}{n+1}=(1+\frac{1}{n^{2}-1})^{n}\frac{n}{n+1}[/tex] now using bernuli inequality we get:

    [tex]\frac{t_n-_1}{t_n}=\frac{(1+\frac{1}{n-1})^{n}}{(1+\frac{1}{n})^{n+1}}=.....=(\frac{n^{2}}{n^{2}-1})^{n}*\frac{n}{n+1}=
    (\frac{n^2-1+1}{n^{2}-1})^{n}\frac{n}{n+1}=
    (1+\frac{1}{n^{2}-1})^{n}\frac{n}{n+1}>(1+\frac{n}{n^{2}-1})\frac{n}{n+1}>(1+\frac{n}{n^{2}})\frac{n}{n+1}=1[/tex]

    hence [tex]t_n-_1>t_n[/tex] which means that the sequence is decreasing...


    Try to show some work of yours, for people here won't do your homework, and this way you may get more replies.


    P.S. This looks more like calculus..lol...where does real analysis come into play here????!!!
     
    Last edited: Mar 12, 2008
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