Series of fractions

Note in the third step

$\large{\frac{3}{2}+(\frac{4}{4}-\frac{3}{4})+(\frac{7}{8}-\frac{4}{8})+...}$

We need first to prove that the series converges. For that we use the ratio test. But before that, first observe that the numerators are in a Fibonacci sequence with starting terms 3 and 4. Hence, if a and b are two consecutive numerators then $b<a<2b$ . Let us denote the numerator sequence as ${ \left( { F }_{ n } \right) }_{ n\ge 1 }$ and let ${ R }_{ n }=\frac { { F }_{ n+1 } }{ { F }_{ n } } ,\forall n\in N$ . Then we have ${ F }_{ n+1 }={ F }_{ n }+{ F }_{ n-1 }\\$ and ${ R }_{ n }=1+\frac { 1 }{ { R }_{ n-1 } }$ . We shall show that this ratio sequence goes to the Golden Ratio $\phi$ given by: $\phi =1+\frac { 1 }{ \phi }$ . We see that: $\left| { R }_{ n }-\phi \right| =\left| \left( 1+\frac { 1 }{ { R }_{ n-1 } } \right) -\left( 1+\frac { 1 }{ \phi } \right) \right| \\ =\left| \frac { 1 }{ { R }_{ n-1 } } -\frac { 1 }{ \phi } \right| \\ =\left| \frac { \phi -{ R }_{ n-1 } }{ \phi { R }_{ n-1 } } \right| \\ \le \left( \frac { 1 }{ \phi } \right) \left| \phi -{ R }_{ n-1 } \right|\\ \le { \left( \frac { 1 }{ \phi } \right) }^{ n-1 }\left| { R }_{ 1 }-\phi \right|$ Which clearly shows that $\left( { R }_{ n } \right) \longrightarrow \phi$ Now we look at the given series and apply the ratio test on it. If we denote the n-th term of the series as ${ x }_{ n }$ then we find that $\frac { { x }_{ n+1 } }{ { x }_{ n } } =\frac { { F }_{ n+1 } }{ { 2F }_{ n } }$ . Now the RHS obviously has limit $\frac { \phi }{ 2 } \approx \frac { 1.618 }{ 2 } \approx 0.809<1$ . Hence ratio test is satisfied and the series converges. The rest of the proof is just same as what Tanishq provided.