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Prove some properties of the sequence ean = an + ebn

by RoRi

Let \{ a_n \} and \{ b_n \} be two sequences that satisfy the relationship

    \[ e^{a_n} = a_n + e^{b_n} \]

for all n.

  1. Prove that if a_n > 0 for all n then b_n > 0 for all n.
  2. Prove that if a_n > 0 for all n and if \sum a_n is convergent, then

        \[ \sum \frac{b_n}{a_n} \qquad \text{converges}. \]

  1. Proof. Assume a_n > 0 for all n. From the given relation between the a_n and b_n we have

        \[ e^{a_n} = a_n + e^{b_n} \quad \implies \quad e^{b_n} = e^{a_n} - a_n. \]

    Now, we claim e^{a_n} - a_n > 1 for all n. To see this, consider the function

        \[ f(x) = e^x - x \qquad \implies \qquad f'(x) = e^x - 1. \]

    Since f'(x) > 0 for all x > 0 we have f(x) is increasing for all x > 0. Since f(0) = 1 we must have f(x) > 1 for all x > 0. Since a_n > 0 by assumption we then have

        \[ e^{a_n} - a_n > 1 \qquad \text{for all } n. \]

    But, this implies

        \[ e^{b_n} > 1 \quad \implies \quad b_n > 0 \]

    for all n. \qquad \blacksquare

  2. Proof. To prove \sum \frac{b_n}{a_n} converges we use the limit comparison test with \sum a_n. First, since \sum a_n converges we know

        \[ \lim_{n \to \infty} a_n = 0. \]

    Now, we use the given relation between the a_n and the b_n,

        \[ e^{a_n} = a_n + e^{b_n} \quad \implies \quad b_n = \log \left( e^{a_n} - a_n \right) = \log \left( 1 + \left( e^{a_n} - 1 - a_n \right) \right). \]

    Then, use the expansion of the exponential,

        \[ e^{a_n} = 1 + a_n + \frac{a_n^2}{2} + \cdots, \quad \implies \quad e^{a_n} - 1 - a_n = \frac{a_n^2}{2} + \frac{a_n^3}{6} + \cdots. \]

    This gives us

        \[ b_n = \frac{a_n^2}{2} + \frac{a_n^3}{6} + \cdots. \]

    Therefore,

        \begin{align*} \lim_{n \to \infty} \frac{\frac{b_n}{a_n}}{a_n} &= \lim_{n \to \infty} \frac{b_n}{a_n^2} \\[9pt] &= \lim_{n \to \infty} \frac{ \frac{a_n^2}{2} + \frac{a_n^3}{6} + \cdots }{a_n^2} \\[9pt] &= \frac{1}{2} + \lim_{n \to \infty} \left( \frac{a_n}{6} + \cdots \right) \\[9pt] &= \frac{1}{2}. &(\lim_{n \to \infty} a_n = 0.) \end{align*}

    Therefore, by the limit comparison test the series \sum \frac{b_n}{a_n} and \sum a_n either both converge or both diverge. Since \sum a_n converges by hypothesis, we have established the convergence of \sum \frac{b_n}{a_n}. \qquad\blacksquare

8 comments

    • Mohammad Azad says:

      I guess you assumed that there is a differentiable function f which satisfies all the conditions of L’Hopital’s rule and f(n)=a_n for all n? Oh and it seems you assumed that f'(x)=1 for all sufficiently large x?

  2. Andres Tellez says:

    Your proof is almost right. From the Taylor series you can deduce that the series of (e^bn-1) converges by the limit comparison test. But from part (a), we know that bn0 (because an>0, then bn>0). This shows that the series of bn converges.

  3. tom says:

    Apparently my comment didn’t post. There is a problem where you used the Taylor series in the comparison – it should be the log of the Taylor series. But more troubling to me is this: if a(n) converges, then a(n) becomes arbitrarily small. So e^a(n) —> e^b(n) because the order of magnitude of e^x is much greater then x. (x=a(n)).

    So e^a(n)/e^b(n) —> 1 implying a(n) —-> b(n). So if limit b(n)/a(n) —> 1 which ≠ 0 how can this converge?

  4. tom says:

    In the comparison test, b(n) should be the log of a Taylor series. Would this work? e^a(n) = a(n) + e^b(n) >e^b(n) (since a(n)>0) implying a(n)>b(n) (since b(n)>0) so b(n) converges.

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