How would you show that the following series is convergent: $$\sum_{n=1}^\infty \left(\left(n+\frac12\right)\ln\left(1+\frac1n\right) - 1 \right)$$

Here is what I have tried so far:

  • I can show that the terms tend to zero, but have been unsuccessful in showing that the series is Cauchy as this seems to amount to just showing that the series as a whole converges (i.e. it does no to simplify the problem)
  • The ratio test does not give an answer
  • The integral test might work, but I have been unable to evaluate the integral
  • Wolfram Alpha suggests the comparison test, so I have tried the following:
    • Comparison with $\frac1{n^2}$ (plotting the graphs, I know that this is a 'correct' comparison) but I cannot get the answer out)
    • Comparison by using the inequality $\ln(1+\frac1n) \le \frac1n$ but this only shows that the series is less than the divergent harmonic series, so has not helped
  • Taylor series expansion for log... though this initially seemed promising, it got a bit messy, and crucially, I wanted to avoid this method as this has not been covered in our analysis course thus far, so the question should not require it
  • I have also tried rewriting it by collecting together the log terms in the sum to get that the series sum is equal to $\lim\limits_{n\to \infty}\ln(\frac{n^n e^{-n}}{n!})$, which seems reminiscent of Stirling's approximation - however, our lecturer said that this could be used to show that Stirling's approximation converges (I can see how to do this part), so I would rather avoid a proof that directly makes use of Stirling's approximation (the implication seems to be that there is a more 'elementary' way to show convergence).

I wanted to work out the answer myself, but have got to the point where I feel I am staring at the question but am unable to make much progress. For this reason, with any solutions, could you possibly include the 'steps'/'clues' that led you to the solution, and how similar questions could be approached.



Initially I was a bit wary of using anything involving Taylor polynomials as it is not something that has been remotely touched on in lectures (only the 'obvious' properties of functions such as log have been used/manipulated in inequalities), but now that I think about it, log is (usually) just defined in terms of its series expansion (correct me if I'm wrong here... I know it is also sometimes defined in terms of an integral). Hence, I feel that answers involving Taylor polynomial are good too... nevertheless, given the fact that none of this has been used in any of the other questions or examples from lectures, I am still wondering if there is another way of proving convergence just using inequality manipulation and only 'basic' logarithm inequalities.

... Also, whilst I am very grateful for all answers, I am interested to see how the problem may be solved in different ways so as to give me more tools examples to help solve further problems

  • 1
    $\begingroup$ A Taylor expansion of $\log (1+x)$ of sufficiently high order ($3$) suffices. $\endgroup$ – Daniel Fischer Dec 7 '16 at 14:50
  • $\begingroup$ We don't need Taylor series, only Taylor polynomials (and corresponding remainder terms) of relatively low order. If you're not comfortable with that, what can one use about the logarithm? Which definition of $\log$ is in use, are there fundamental inequalities one can employ? $\endgroup$ – Daniel Fischer Dec 7 '16 at 15:00

Find an equivalent of the general term. We'll have to expand the log at order $3$: \begin{align}\Bigl(n+\frac12\Bigr)\ln\Bigl(1+\frac1n\Bigr)-1&=\Bigl(n+\frac12\Bigr)\Bigl(\frac1n-\frac 1{2n^2}+\frac1{3n^3}+o\Bigl(\frac1{n^3}\Bigr)\Bigr)-1\\&=1-\frac1{2n}+\frac1{3n^2}+o\Bigl(\frac1{n^2}\Bigr)+\frac1{2n}-\frac 1{4n^2}+\frac1{6n^3}+o\Bigl(\frac1{n^3}\Bigr)-1\\ &=\frac 1{12n^2}+o\Bigl(\frac1{n^2}\Bigr) \end{align} Thus $\;u_n\sim_\infty\dfrac 1{12n^2}$, which converges.

  • $\begingroup$ I would have written the initial error term as $O(1/n^4)$ which would give a final error of $O(1/n^3)$. $\endgroup$ – marty cohen Dec 7 '16 at 15:30
  • $\begingroup$ Personally, in these questions, I prefer to use the o notation, as it is involved in the definition of equivalence. $\endgroup$ – Bernard Dec 7 '16 at 15:34
  • $\begingroup$ Thanks for the answer... I don't know how this eluded me for so long, especially as I had the first line written down at one point $\endgroup$ – John Don Dec 7 '16 at 15:40
  • $\begingroup$ As usual, asymptotic computation yields a straightforward solution. $\endgroup$ – Gabriel Romon Dec 7 '16 at 17:51
  • $\begingroup$ @Bernard: I prefer big-o when possible because, as in this case, it more accurately reflects the true error of the approximation. $\endgroup$ – marty cohen Dec 7 '16 at 22:04

By L'Hopital's Rule, one has \begin{eqnarray} &&\lim_{n\to\infty}\frac{(n+\frac12)\ln(1+\frac1n)-1}{\frac1{n^2}}\\ &=&\lim_{x\to0}\frac{(\frac1x+\frac12)\ln(1+x)-1}{x^2}\\ &=&\lim_{x\to0}\frac{(x+2)\ln(1+x)-2x}{2x^3}\\ &=&\frac1{12}. \end{eqnarray} Noting that $\sum_{n=1}^\infty\frac1{n^2}$ converges and so $\sum_{n=1}^\infty\big[(n+\frac12)\ln(1+\frac1n)-1\big]$ converges.

Update: Without using L'Hopital's Rule. Note $$ \ln(1+x)\le x-\frac{x^2}{2}+\frac13x^3, x>0. $$ Using this, one has \begin{eqnarray} 0\le(n+\frac12)\ln(1+\frac1n)-1&\le&(n+\frac12)(\frac1n-\frac{1}{2n^2}+\frac1{3n^3})-1\\ &=&\frac{1}{12}\frac1{n^2}+\frac{1}{6n^3} \end{eqnarray} So $\sum_{n=1}^\infty\frac1{n^2}$ converges.

  • $\begingroup$ This one is the comparison test. $\endgroup$ – xpaul Dec 7 '16 at 15:54
  • $\begingroup$ One of my lecturer's mentioned that there are various other tests for convergence that 'elaborate' on the Comparison test when the simple ratio or comparison tests do not give a result... I have tried searching for a few but they don't seem to apply directly in this case.... I was wondering if perhaps another test (similar to the comparison test) could be used to compare the series to $\frac1{n^2}$ without the need to use l'hopital's rule $\endgroup$ – John Don Dec 7 '16 at 16:58

We can prove that the sum converges by proving that the limit of the partial sums converges. We can do that by simply solving the sum!

Define: $$ S(N) = \sum_{n = 1}^{N-1} \left[ \left(n + \frac{1}{2}\right)\ln\left(1 + \frac{1}{n}\right) - 1\right]\, . $$ (We define the upper limit as $N-1$ rather than $N$ because it makes the math "prettier" later.) I can rearrange this as follows: \begin{align} S(N) &= \sum_{n = 1}^{N-1} \left[ \left(n + \frac{1}{2}\right)\ln\left(\frac{n+1}{n}\right)\right] - (N-1)\\ &= \sum_{n = 1}^{N-1} \left[ \left(n + 1 + \frac{1}{2}\right)\ln\left(n+1\right) - \left(n + \frac{1}{2}\right)\ln\left(n\right) - \ln(n+1) \right] - (N-1)\\ &= \sum_{n = 1}^{N-1} \left[ \left(n + 1 + \frac{1}{2}\right)\ln\left(n+1\right) - \left(n + \frac{1}{2}\right)\ln\left(n\right) \right] - \ln\left(N!\right)- (N-1) \end{align} Note that the argument of the remaining sum can be written as $$ \sum_{n = 1}^{N-1} \left[ \left(n + 1 + \frac{1}{2}\right)\ln\left(n+1\right) - \left(n + \frac{1}{2}\right)\ln\left(n\right) \right] = \sum_{n = 1}^{N-1}\left[f(n+1) - f(n)\right]\, , $$ where $f(x) = \left(x + \frac{1}{2}\right)\ln(x)$. Thus this sum telescopes, leaving: $$ \sum_{n = 1}^{N-1} \left[ \left(n + 1 + \frac{1}{2}\right)\ln\left(n+1\right) - \left(n + \frac{1}{2}\right)\ln\left(n\right) \right] = f(N) - f(1) = \left(N+\frac{1}{2}\right)\ln(N) $$ Putting it all together: $$ S(N) = \left(N+\frac{1}{2}\right)\ln(N)- \ln\left(N!\right)- (N-1) $$ We want to take the limit of $S(N)$ as $N\rightarrow\infty$, so we invoke the Stirling approximation for the factorial: \begin{align} \lim_{N\rightarrow\infty}S(N)&= \lim_{N\rightarrow\infty}\left\{ \left(N+\frac{1}{2}\right)\ln(N)- \ln\left(N^N e^{-N} \sqrt{2\pi N}\,\right)- (N-1)\right\}\\ &= 1 - \ln\left(\sqrt{2\pi}\right)\approx 0.08106 \end{align}

(In the last step above, all the terms inside the limit involving $N$ cancel each other out, so we're just left with the constant.)

Numerically evaluating the sum (in Mathematica) gives the same result.

  • $\begingroup$ Nice work noticing the telescoping. I think you were the only one to get the actual sum. $\endgroup$ – marty cohen Dec 7 '16 at 22:07

The integral test works. Define $$ I:= \int \left(x+\frac{1}{2} \right)\ln\left( 1+ \frac{1}{x}\right) -1 \;\mathrm dx =\int \left(x+\frac{1}{2} \right)\ln\left( 1+ \frac{1}{x}\right) \;\mathrm dx - \int1 \; \mathrm d x .$$ Integration by parts with $u'=x+\frac{1}{2}$ and $v=\ln\left( 1+ \frac{1}{x}\right)$ gives $$ I = \frac{1}{2}\left(x^2 +x \right)\ln\left( 1+ \frac{1}{x}\right) - \frac{1}{2}\int_0^\infty (x^2+x)\cdot\frac{-1}{x^2+x} \; \mathrm d x -\int_0^\infty 1 \; \mathrm d x$$ and hence $$I=\frac{1}{2}\left(x^2 +x \right)\ln\left( 1+ \frac{1}{x}\right)- \frac{1}{2}x.$$ Using a Taylor polynomial for $\ln$ gives $$\lim_{x\to {+\infty} }\frac{1}{2}\left(x^2 +x \right)\ln\left( 1+ \frac{1}{x}\right)- \frac{1}{2}x= \frac{1}{2}$$ and therefore $$\int_0^{+\infty} \left(x+\frac{1}{2} \right)\ln\left( 1+ \frac{1}{x}\right) -1 \;\mathrm dx =\frac{3}{4} - \ln2 < + \infty$$ as desired.

  • $\begingroup$ How would you show that the integrand is a decreasing function? I think I can see a way, but it is a bit messy... am I missing something obvious? $\endgroup$ – John Don Dec 7 '16 at 15:36
  • 1
    $\begingroup$ The derivative of the integrand is $-(x^3+x^2/2)/(x+1)+\log (1+1/x)=$ $-x^2+x^2/(2x+2)+\log (1+1/x).$ For $x>1$ this is less than $-(x^2-x/2-1/x)$ which is negative ..(We have $x^2/(2x+2)<x^2/(2x)=x/2$ and $\log (1+1/x)< 1/x.)$ $\endgroup$ – DanielWainfleet Dec 8 '16 at 1:16

Let $t(n) =(n+\frac12)\ln(1+\frac1n)-1 $ and $S(N) =\sum_{n=1}^N t(n) $.

I will show that $t(n) =\frac1{n^2}\int_0^{1/2}\frac{y(2y-1)\,dy}{1-((1/2-y)/n)^2} $ so that $0 > t(n) > -\frac1{6n^2} $ which implies that $S(N)$ converges.

$\begin{array}\\ (n+\frac12)\ln(1+\frac1n) &=(n+\frac12)\int_0^{1/n}\frac{dx}{1+x}\\ &=(n+\frac12)\frac1{n}\int_0^{1}\frac{dx}{1+x/n}\\ &=(1+\frac1{2n})\int_0^{1}\frac{dx}{1+x/n}\\ \end{array} $


$\begin{array}\\ t(n) &=(n+\frac12)\ln(1+\frac1n)-1\\ &=(1+\frac1{2n})\int_0^{1}\frac{dx}{1+x/n}-1\\ &=\int_0^{1}\frac{dx}{1+x/n}(1+\frac1{2n}-(1+\frac{x}{n}))\\ &=\int_0^{1}\frac{dx}{1+x/n}(\frac1{2n}-\frac{x}{n})\\ &=\frac1{n}\int_0^{1}\frac{dx}{1+x/n}(\frac1{2}-x)\\ &=\frac1{n}\left(\int_0^{1/2}\frac{dx}{1+x/n}(\frac1{2}-x)+\int_{1/2}^{1}\frac{dx}{1+x/n}(\frac1{2}-x)\right)\\ &=\frac1{n}\left(\int_0^{1/2}\frac{y\,dy}{1+(1/2-y)/n}-\int_{0}^{1/2}\frac{y\,dy}{1+(y-1/2)/n}\right)\\ &=\frac1{n}\int_0^{1/2}y\,dy\left(\frac1{1+(1/2-y)/n}-\frac1{1+(y-1/2)/n}\right)\\ &=\frac1{n}\int_0^{1/2}y\,dy\frac{(1+(y-1/2)/n)-(1+(1/2-y)/n)}{(1+(1/2-y)/n)(1+(y-1/2)/n)}\\ &=\frac1{n^2}\int_0^{1/2}y\,dy\frac{(y-1/2)-(1/2-y)}{(1+(1/2-y)/n)(1-(1/2-y)/n)}\\ &=\frac1{n^2}\int_0^{1/2}\frac{y(2y-1)\,dy}{1-((1/2-y)/n)^2}\\ \end{array} $

Since, if $0\le y \le \frac12$, $-\frac18 \le y(2y-1) \le 0$ (because $y(2y-1) =\frac18( (4 y - 1)^2 - 1) $) and $0 \le ((1/2-y)/n)^2 \le \frac1{4n^2} $, $t(n) \le 0 $ and, since $n \ge 1$, $t(n) \gt -\frac1{8n^2}\frac1{1-1/(4n^2)} \ge -\frac1{8n^2}\frac43 \ge -\frac1{6n^2} $.

Therefore $\sum_{n=1}^{\infty} t(n)$ converges.


Let $x=\frac {1}{2n+1}.$ Because typing.

$$\text {We have }\quad 1+\frac {1}{n}=\frac {1+x}{1-x}.$$ $$\text {So }\quad \log (1+\frac {1}{n})=$$ $$=\log (1+x)-\log (1-x)=(x-x^2/2+x^3/3-...)-(-x-x^2/2-x^3/3-...)=$$ $$=2x (1+x^2/3+x^4/5+...).$$ $$\text {So }\quad (n+\frac {1}{2})\log (1+\frac {1}{n})-1=\frac {1}{2x}[2x(1+x^2/3+x^4/5+...)]-1=$$ $$= x^2/3+x^4/5+...$$ which is positive but less than $(x^2/3)(1+x^2+x^4+...)= (x^2/3)/(1-x^2),$ which is less than $(x^2/3)\cdot 2,$ which is equal to $$\frac {2}{3(2n+1)^2}.$$


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