How to prove $(1-\frac1{36})^{25}\lt\frac12$?

Question

How to prove the inequality?

$(1-\frac1{36})^{25}\lt\frac12$

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I'm in trouble.

Thank you very much for your help

Answer 1

The series $$(1-x)^{-25}=1+25 x+325 x^2 + 2925 x^3+20\,475 x^4+\ldots$$ has all coefficients positive. It follows that $$\left(1-{1\over36}\right)^{-25}> 1+{25\over 36}+{325\over 36^2}+{2916\over 36^3}=1+{25\over36}+{325+81\over 36^2}={1301\over648}>2\ .$$

Answer 2

Without using either large numbers, large amount of terms or evaluating the value as decimal, I can't think of any way to solve this, here's one way that involves large numbers

$$ \begin{array}{rl} &\left(1-\frac1{36}\right)^{25}\\ =&\left(\frac{35}{36}\right)^{25}\\ =&\frac{35^{25}}{36^{25}}\\ =&\frac{399669593472470313551127910614013671875}{808281277464764060643139600456536293376} \end{array} $$

Since if $\frac ab < \frac12$ then $2a < b$

$$ \begin{array}{rl} &399669593472470313551127910614013671875\times 2\\ =&799339186944940627102255821228027343750 \end{array} $$

Now we have 2 integers, we can compare these $$ \begin{array}{c} 799339186944940627102255821228027343750\\ 808281277464764060643139600456536293376 \end{array} $$ Obviously the lower number is the largest, so the inequality holds.

Answer 3

We have that $\displaystyle\left(1-\frac{1}{36}\right)^{25}=\sum_{r=0}^{25} \binom{25}{r}(-1)^{r}\frac{1}{36^r}=1-\frac{25}{36}+\frac{25\cdot24}{2\cdot36^2}-\frac{25\cdot24\cdot23}{6\cdot36^3}+\frac{25\cdot24\cdot23\cdot22}{24\cdot36^4}-\cdots$.

Since the sum is alternating and has terms that are decreasing in absolute value, its value is less than

$\displaystyle1-\frac{25}{36}+\frac{25\cdot24}{2\cdot36^2}-\frac{25\cdot24\cdot23}{6\cdot36^3}+\frac{25\cdot24\cdot23\cdot22}{24\cdot36^4}=\frac{29}{54}-\frac{25\cdot23\cdot122}{36^4}<\frac{1}{2}$ since

$27(25)(23)(122)>27(24)(23)(120)=9(12^{2})(3)(2)(23)(10)=36^{2}(1380)>36^2(1296)=36^{4}$

$\implies\displaystyle\frac{25\cdot23\cdot122}{36^{4}}>\frac{1}{27}.$

Answer 4

Note that $\left(1-\frac1{36}\right)^{36}\approx e^{-1}$, but more specifically $\left(1-\frac1{36}\right)^{36}\lt e^{-1}$ (because the function $\left(1-\frac1x\right)^x$ is increasing for $x\gt 1$ and its limit as $x\to\infty$ is $e^{-1}$). Thus, $\left(1-\frac1{36}\right)^{25} =\left(\left(1-\frac1{36}\right)^{36}\right)^{\frac{25}{36}}\lt e^{-\frac{25}{36}}$. But we have $e^{-\frac{25}{36}}\lt\frac12$ because $e^{\frac{25}{36}}\gt 2$ because $\frac{25}{36}\gt\ln 2$. This last equation can easily be checked with a small table of constants.

Answer 5

Take logarithms

Left Hand Side $= 25(\log 35 - \log 36) = -.30586...$;
Right Hand Side $= -\log 2= -.30102...$

Left Hand Side is less than the Right Hand Side

Answer 6

Using logarithms, but without the aid of a calculator, you need to show $$25\ln\left(\frac{35}{36}\right)<-\ln(2)$$ which is the same as $$25\ln(7)+25\ln(5)-50\ln(3)-49\ln(2)<0$$

Now at this point, you go old-school and have $\ln$ of various prime numbers memorized:

$$25(1.9549\ldots)+25(1.6094\ldots)-50\ln(1.0986\ldots)-49(0.6931\ldots)$$

which is a net negative.

Answer 7

Note that the inequality holds iff $35/36<(1/2)^{1/25}.$ The continued fraction for the latter number starts out $[0;1,35,1,1,3,11,...]$ and the even convergents are all less than $(1/2)^{1/25}.$ The first even convergent (i.e. the second convergent) is $0+1/(1+1/35)=35/36,$ so this shows the inequality, provided one is allowed to use the computation of the continued fraction. The next even convergent, the fourth, is $71/73$, and its 25th power is again less than $1/2$ and so on.

Edit: Note that I had to use decimal evaluations to obtain the first few terms of the continued fraction for $(1/2)^{1/25}.$ In a way that is a drawback, since just finding the decimal $(1/2)^{1/25} \approx 0.97265$ already finishes the problem, since $35/36 \approx 0.972222.$ However maybe there is a way to get the continued fraction, at least the first two entries, without resorting to decimals. Things here are more complicated than in the easier case of squareroots, in which rationalizing works at the steps. It was easy enough getting to $[0;1]$ for the (trivial) first convergent, but trying to actually prove it continues $[0;1,35]$ didn't seem to give any nice provable calculation because of the presence of 25th roots.

However it did seem like an interesting coincidence that the second convergent came out exactly the $35/36$ of the posted inequality.

Answer 8

Using logarithms, as already said, you have to show that $$25\ln\left(\frac{35}{36}\right)<-\ln(2)$$ A very good Taylor expansion of logarithms is given by $$\ln\left(\frac{1+x}{1-x}\right)=2 \Big(\frac{x}{1}+\frac{x^3}{3}+\frac{x^5}{5}+....\Big)$$ defining $x$ from $\frac{1+x}{1-x}=\frac{35}{36}$ gives $x=-\frac{1}{71}$ which is quite small and the first term should be sufficient. So,$$25\ln\left(\frac{35}{36}\right) \simeq -\frac{50}{71} \simeq -0.704225$$ which is already smaller than $-\ln(2) \simeq -0.693147$. If we use a second term (remember that $x <0$ and that we raise it to the third power), then the difference with still increase since $$25\ln\left(\frac{35}{36}\right) \simeq -50\Big(\frac{1}{71}+ \frac{1}{3 \times 71^3}\Big)=-\frac{756200}{1073733}\simeq -0.704272$$

Answer 9

All answers so far have been approximations of the value or its logarithm. As such, none seem particularly appealing. However, using the binomial theorem seems most appealing to me since it can be done with the least computation, and a good bit of reused computation. Therefore, this computation can reasonably be performed by hand, and this is what I think the question is asking for since given a calculator it is not too hard simply to compute $$ \left(\frac{35}{36}\right)^{25}=0.49446845 $$

The approach in this answer has already been given in the answer by user84413, and I think it is the simplest to compute since each binomial coefficient is quite easy to compute from the last and the same is true for powers of $\frac1{36}$.

It should also be pointed out that any error from rounding in one term is multiplied by a factor of less than $1$ before being passed on to the next term, so each term should be accurate to the last digit, so even though the computation below is only carried out to $4$ decimal places, the error should be less than $0.00005\times4=0.0002$. (In fact, the computation turns out to be correct to the last decimal place.) $$ \begin{align} \left(1-\frac1{36}\right)^{25} &\le1-\frac{\binom{25}{1}}{36^1}+\frac{\binom{25}{2}}{36^2}-\frac{\binom{25}{3}}{36^3}+\frac{\binom{25}{4}}{36^4}\\ &=1-\frac{25}{36}+\color{#808080}{\frac{25}{36}}\frac{24}{2\cdot36}-\color{#808080}{\frac{25}{36}\frac{24}{2\cdot36}}\frac{23}{3\cdot36}+\color{#808080}{\frac{25}{36}\frac{24}{2\cdot36}\frac{23}{3\cdot36}}\frac{22}{4\cdot36}\\[6pt] &=1-0.6944+0.2315-0.0493+0.0075\\[9pt] &=0.4953\\[9pt] &\lt0.5000 \end{align} $$

Answer 10

Here I present a basic proof only using the binomial expansion theorem without using the property of $(1-\frac{1}{x})^x$. Note that the inequality is equivalent to the following inequality $$ \left(\frac{36}{35}\right)^{25}> 2.$$ By the binomial expansion theorem, we have \begin{eqnarray} \left(\frac{36}{35}\right)^{25}&=&\left(1+\frac{1}{35}\right)^{25}\\ &>&1+\binom{25}{1}\frac{1}{35}+\binom{25}{2}\frac{1}{35^2}+\binom{25}{3}\frac{1}{35}\\ &=&1+\frac{25}{35}+\frac{25\cdot 24}{2}\frac{1}{35^2}+\frac{25\cdot24\cdot23}{3\cdot 2}\frac{1}{35^3}\\ &=&1+\frac{5}{7}+\frac{12}{49}+\frac{92}{1715}\\ &=&1+\frac{1737}{1715}\\ &>&2. \end{eqnarray} Done.

Answer 11

If you are willing to use that $\ln(1 - x) < -x$ (which is immediate from the power series for $\ln(1 - x)$) then it can be done quite quickly. Taking logs your inequality is equivalent to $$25 \ln \bigg(1 - {1 \over 36}\bigg) < \ln (1/2)$$ Using $\ln(1 - x) < -x$ it suffices to show that $$-{25 \over 36} < \ln (1/2)$$ Equivalently, $$\ln 2 < {25 \over 36}$$ Note that $\ln 2 = .6931...$ and ${25 \over 36} = .6944..$

So the inequality does hold.