Mind your 3's and 2's

Note that $\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\cdots\right)\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+\cdots\right)$ covers everything we want from distributive property. We evaluate this using the infinite geometric sum formula to get $(1)(\frac{1}{2})=\dfrac{1}{2}$ so the correct answer is $\boxed{3}$ .

The answer to this question is incorrect (they put $5$ ).

The problem can be expressed as finding the product of two infinite geometric sequences. $\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\dfrac{1}{2^i3^j} =\sum_{i=1}^{\infty}\frac{1}{2^i}\times\sum_{j=1}^{\infty}\frac{1}{3^i}$ We can find the sums of these two series, $M$ and $N\text{,}$ individually and multiply them together. Let's start by solving for $M\text{.}$

$M=\sum_{i=1}^{\infty}\frac{1}{2^i}$ This is a geometric sequence with first term $\frac{1}{2}$ and ratio $\frac{1}{2}$ , so the sum of this sequence is $\dfrac{\frac{1}{2}}{1-\frac{1}{2}}=\dfrac{\frac{1}{2}}{\frac{1}{2}}=\textbf{1}$

Now we can solve for $N\text{.}$ $N=\sum_{j=1}^{\infty}\frac{1}{3^j}$ This is a geometric sequence with first term $\frac{1}{3}$ and ratio $\frac{1}{3}$ , so the sum of this sequence is $\dfrac{\frac{1}{3}}{1-\frac{1}{3}}=\dfrac{\frac{1}{3}}{\frac{2}{3}}=\frac{\textbf{1}}{\textbf{2}}$

Multiplying $M$ and $N$ gives us the desired answer. $1\times\frac{1}{2}=\frac{1}{2}\text{,}$ so $A=1$ and $B=2\text{.}$ $A+B=\boxed{3}$

We can write the sum of the reciprocals as follows: $\sum_{n=1}^\infty{\sum_{i=1}^\infty{\frac{1}{2^{i} \times 3^{n}}}}$ We can extract $\frac{1}{3^{n}}$ from the inner sum (since it is constant relative to i): $\sum_{n=1}^\infty{\frac{1}{3^{n}} \times \sum_{i=1}^\infty{\frac{1}{2^{i}}}}=\sum_{n=1}^\infty{\frac{1}{3^{n}} \times 1}=\sum_{n=1}^\infty{\frac{1}{3^{n}}}=\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}=\frac{A}{B}$ Thus $A+B=\boxed{3}$ .

The sum is:

$\begin{aligned} S & = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac 1{2^m3^n} = \sum_{m=1}^\infty \frac 1{2^m} \sum_{n=1}^\infty \frac 1{3^n} = \frac 12 \left( \frac 1{1-\frac 12} \right) \cdot \frac 13 \left( \frac 1{1-\frac 13} \right) = \frac 12 \cdot 2 \cdot \frac 13 \cdot \frac 32 = \frac 12 \end{aligned}$

$\implies A+B = \boxed{3}$

First, we keep increasing the power of $2$ and keep $3$ the same. We get

$\Rightarrow 2.3 + 2^{2}.3 + 2^{3}.3 + \dots$ . Now, the reciprocal of this is equal to

$\Rightarrow \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{24} + \dots$

This looks like an infinite geometric series. The sum of this infinite geometric series is

$\Rightarrow \dfrac{\dfrac{1}{6}}{\dfrac{1}{2}} = \dfrac{1}{3}$ .

Now, we increment the exponent of $3$ by $1$ , making it $3^2$ and then form the series of numbers.

Thus, we get

$\Rightarrow 2.3^{2} + 2^{2}.3^{2} + 2^{3}.3^{2} \dots$

The reciprocal of these numbers is

$\Rightarrow \dfrac{1}{18} + \dfrac{1}{36} + \dfrac{1}{72} + \dots$

The sum of this infinite geometric series is

$\Rightarrow \dfrac{\dfrac{1}{18}}{\dfrac{1}{2}} = \dfrac{1}{9}$

We do this iteratively, that is, increment powers of $3$ by $1$ , now making it $3^3$ and then forming an infinite series and then summing it up.

If we add these infinite sums, we get a series of the form

$\Rightarrow \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \dots$

This is also an infinite geometric series with common ratio $\dfrac{1}{3}$ . (Wow, an infinite series of the sums of infinite series')

We can sum this, to get

$\Rightarrow \dfrac{\dfrac{1}{3}}{\dfrac{2}{3}} = \dfrac{1}{2}$ .

Thus, our answer is $1 + 2 = \boxed{3}$ .

sum of numbers can be -

(1/2)[1/3+1/9+1/27+......]+1/4[1/3+1/9+1/27+......]+1/8[1/3+1/9+1/27.....]+.......

=[1/2+1/4+1/8........][1/3+1/9+1/27.......]

=[(1/2)/(1/2)][(1/3)/(2/3)]

=1/2

=> A + B =1+2=3

The sum is:

$\frac{1}{2\times3}+\frac{1}{2\times3^{2}}+\frac{1}{2\times3^{3}}+\frac{1}{2\times3^{4}}+\frac{1}{2\times3^{5}}+\frac{1}{2\times3^{6}}...$

$+\frac{1}{2^2\times3}+\frac{1}{2^2\times3^{2}}+\frac{1}{2^2\times3^{3}}+\frac{1}{2^2\times3^{4}}+\frac{1}{2^2\times3^{5}}+\frac{1}{2^2\times3^{6}}...$

$+\frac{1}{2^3\times3}+\frac{1}{2^3\times3^{2}}+\frac{1}{2^3\times3^{3}}+\frac{1}{2^3\times3^{4}}+\frac{1}{2^3\times3^{5}}+\frac{1}{2^2\times3^{6}}...$

By some factorization, we find that the sum is:

$(\sum_{p=1}^\alpha \frac{1}{2^p})(\sum_{q=1}^\alpha \frac{1}{3^q})$

Here, $\sum_{p=1}^\alpha \frac{1}{2^p}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$

$\sum_{q=1}^\alpha \frac{1}{3^q}=\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{2}$

So the sum is $1\times\frac{1}{2}=\frac{1}{2}$ ,and the answer is $1+2=\boxed{3}$

The required sum is given by $(1+ \frac{1}{2^{1}} + \frac{1}{2^{2}} + \frac{1}{2^{3}}....) \times (1+ \frac{1}{3^{1}} + \frac{1}{3^{2}} + \frac{1}{3^{3}} ....) -1$ $= \frac{1}{1-\frac{1}{2}} \times \frac{1}{1-\frac{1}{3}} -1$ $= (2 \times \frac{3}{2}) -1$ = $2$

thus, $A+B = 2+1 =3$

Result = ∑m,n (1 -> infinity) 1/(2^m 3^n)= ∑m (1 -> infinity) 1/(2^m){∑n (1 -> infinity)1/(3^n)} =∑m (1 -> infinity)1/(2^m) 1/2=1/2 2 1/2=1/2=A/B A+B=1+2=3

(1/1+ 1/3 + 1/9 + ...) (1/1 + 1/2 + 1/4 + ...) = (1/(2/3))(1/(1/2)) = 3. So A/B=3

Writing as a series, it comes to (1/2)[(1/3)+(1/3)^2+(1/3)^3+.........]+(1/2)^2[(1/3)+(1/3)^2+.........]+(1/2)^3[(1/3)+(1/3)^2+(1/3)^3+....]+......... = [(1/2)+(1/2)^2+(1/2)^3+........][(1/3)+(1/3)^2+(1/3)^3+......]. Both, the terms form infinite G.P. The expression = [(1/2)/{1-(1/2)}][(1/3)/{1-(1/3)}] = 1*[1/2]=1/2.

You can fix 2^m and then change 3^m to get a geometric series with ratio 1/3 Then when youve found the sum to infinty for that geometric series, You can factor out that sum to infinity while changing 2^m now to get a geometric series in 1/2^m now with ratio 1/2.

So the final answer for A/B is the product of the two sum to infinities. using a/(1-r) you get [(1/2)/[1-(1/2)]] $*$ [(1/3)/[1-(1/3)]]=1*1/2=1/2

so A+B= $1+2= \boxed{3}$

This is a double series problem. First fix the value of m and keep on increasing the value of n. This will give you geometric infinite sequences and their sum will be $1/(1-r)(a1+a2+......)$ . $r = 1/3$ . The other sequence is in the first terms of the sequences which is $a = 1/(2*3)$ and $r = 1/2$ and this sequence's sum is given by $a/(1-r) = 1/6*2 = 1/3$ . Putting this value of $a1+a2+...$ and in the first expression we get its value as $1/2$ . $A=1$ and $B=2$ . $1+2=3$

The sum of reciprocals of all such numbers with prime factorisation $2^{m}3^{a}$ for some fixed positive integer $a$ is $\sum_{i=1}^\infty \frac{1}{2^{i}3^{a}}=\frac{1}{3^{a}}\sum_{i=1}^\infty \frac{1}{2^{i}}$ Let A= $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$

Then 2A= $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$

Thus A= $\sum_{i=1}^\infty\frac{1}{2^i}=2A-A=1$

In summary, given any positive integer $a$ , the sum of reciprocals of integers with prime factorisation $2^{m}3^{a}$ is equal to $\frac{1}{3^a}$ .

So the final answer will be just $\sum_{i=1}^\infty\frac{1}{3^i}$

Using the same method as above, we can easily compute the summation.

Let B= $\frac{1}{3}+\frac{1}{9}+\frac{1}{27}...$

Then 3B= $1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}...$

So 2B=1 and thus the answer should be 1+2=3. I happened to guess the answer 5 but I do not think it is correct. If there is anything wrong could you please comment?