The value of $1+(1+\frac{1}{3})\frac{1}{4}+(1+\frac{1}{3}+\frac{1}{9})\frac{1}{16}+(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27})\frac{1}{64}+\cdots$, is

To find the value of the given infinite series, let us first write down the general term of the series.
The given series is:

$S=1+(1+\frac{1}{3})\frac{1}{4}+(1+\frac{1}{3}+\frac{1}{9})\frac{1}{16}+(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27})\frac{1}{64}+\cdots$

Observe the structure of the terms. The $n$-th term, where $n\ge 1$, can be written as:

$T_n=(1+\frac{1}{3}+\frac{1}{3^2}+\cdots +\frac{1}{3^{n-1}})·\frac{1}{4^{n-1}}$

The sum inside the parentheses is a finite geometric progression (GP) with first term $a=1$, common ratio $r=\frac{1}{3}$, and $n$ terms. Using the formula for the sum of a GP, we have:

$1+\frac{1}{3}+\frac{1}{3^2}$

Substituting this back into the formula for the general term $T_n$:

$T_n=\frac{3}{2}(1-\frac{1}{3^n})\frac{1}{4^{n-1}}$

Let us simplify the second term in the parenthesis:
$3^n·4^{n-1}=3·3^{n-1}·4^{n-1}=3·{12}^{n-1}$
Thus, the general term is:

$T_n=\frac{3}{2}(\frac{1}{4^{n-1}}-\frac{1}{3·{12}^{n-1}})$

Now, the sum of the infinite series $S$ is the sum of $T_n$ from $n=1$ to $\infty$:

$S=\sum _{n=1}^{\infty }{T}_n=\frac{3}{2}\sum _{n=1}^{\infty }(\frac{1}{4^{n-1}}-\frac{1}{3·{12}^{n-1}})$

We can split this into two separate infinite geometric series:

$S=\frac{3}{2}(\sum _{n=1}^{\infty }\frac{1}{4^{n-1}}-\frac{1}{3}\sum _{n=1}^{\infty }\frac{1}{{12}^{n-1}})$

The first infinite GP is $1+\frac{1}{4}+\frac{1}{16}+\cdots$ which has first term $a_1=1$ and common ratio $r_1=\frac{1}{4}$. Its sum is:
$S_1=\frac{1}{1-\frac{1}{4}}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$

The second infinite GP is $1+\frac{1}{12}+\frac{1}{144}+\cdots$ which has first term $a_2=1$ and common ratio $r_2=\frac{1}{12}$. Its sum is:
$S_2=\frac{1}{1-\frac{1}{12}}=\frac{1}{\frac{11}{12}}=\frac{12}{11}$

Now, substitute these sum values back into the expression for $S$:

$S=\frac{3}{2}(\frac{4}{3}-\frac{1}{3}·\frac{12}{11})$

Simplify the term inside the parenthesis:
$\frac{4}{3}-\frac{4}{11}=\frac{44-12}{33}=\frac{32}{33}$

Finally, multiply by $\frac{3}{2}$:

$S=\frac{3}{2}·\frac{32}{33}=\frac{1}{1}·\frac{16}{11}=\frac{16}{11}$

Therefore, the value of the infinite series is 16/11.