This is the definition of a convergent infinite series.

Since this is a geometric series with a rate between $\displaystyle 0$ and $\displaystyle 1$, we can use the following equation to find the sum:

$\displaystyle S_{\infty}=\frac{a_{1}}{1-r}$, where $\displaystyle a_{1}$ is the starting number in the sequence, and $\displaystyle r$ is the common divisor between successive terms in the sequence.  In this sequence, to go from one number to the next, we multiply by $\displaystyle \frac{1}{2}.$  Now, we plug everything into the equation:

$\displaystyle S_{\infty}=\frac{a_{1}}{1-r}=\frac{2}{1-0.5}=\frac{2}{0.5}=4.$

Since this is a geometric series with a rate between $\displaystyle 0$ and $\displaystyle -1$, we can use the following equation to find the sum:

$\displaystyle S_{\infty}=\frac{a_{1}}{1-r}$, where $\displaystyle a_{1}$ is the starting number in the sequence, and $\displaystyle r$ is the common divisor between successive terms in the sequence.  In this sequence, to go from one number to the next, we multiply by $\displaystyle \frac{-1}{4}.$  Now, we plug everything into the equation:

$\displaystyle S_{\infty}=\frac{a_{1}}{1-r}=\frac{-240}{1--0.25}=\frac{-240}{1.25}=-192.$

For the sum of an infinite series, we have the following formula:

$\displaystyle S_{\infty}=\frac{a_{1}}{1-r}$, where $\displaystyle a_{1}$ is the first term in the series and $\displaystyle r$ is the rate at which our series is changing between consecutive numbers in the series.  Plugging all of the relevant information for this series, we get:

$\displaystyle S_{\infty}=\frac{80}{1-0.5}=160.$

For the sum of an infinite series, we have the following formula:

$\displaystyle S_{\infty}=\frac{a_{1}}{1-r}$, where $\displaystyle a_{1}$ is the first term in the series and $\displaystyle r$ is the rate at which our series is changing between consecutive numbers in the series.  Plugging all of the relevant information for this series, we get:

$\displaystyle S_{\infty}=\frac{-500}{1--0.2}=\frac{-1250}{3}=-416.666.$

$\displaystyle f(x)=\frac{4x}{x+2}$

This function can be easily written as a power series using the formula for a convergent geometric series. 

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$\displaystyle \sum_{n=1}^{\infty}ar^n =\frac{a}{1-r}$ 

For any $\displaystyle \left | r\right |< 1$

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First let's make some modifications to the function so we can compare it to the form of a convergent geometric series: 

$\displaystyle f(x)=\frac{4x}{x+2}=x\left(\frac{4}{2-(-x)} \right )=x\left(\frac{2}{1-\left(-\frac{x}{2} \right )} \right )$

Notice if we take $\displaystyle r =- \frac{x}{2}$ and $\displaystyle a =2$ we can write $\displaystyle f(x)$ in the form, 

 $\displaystyle f(x)=x\sum_{x=1}^{\infty}2\left(-\frac{x}{2} \right )^n$

We can find the radius of convergence by applying the condition  $\displaystyle \left | r\right |< 1$.

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Finding Radius of Convergence. 

 $\displaystyle \left | \: -\frac{x}{2}\right |< 1$

 Case 1

$\displaystyle - \frac{x}{2}< 1\: \: \: \: \: \: \Rightarrow \: \: \: \: \: \: \frac{x}{2}>-1\: \: \: \: \: \: \:\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, x>-2$

 Case 2

$\displaystyle -\frac{x}{2}>-1\: \: \, \: \: \Rightarrow\, \, \, \, \, \, \, \, \frac{x}{2}< 1\, \, \, \, \, \, \Rightarrow\, \, \, \, \, \, \, \, \, x< 2$      

Combing both cases gives the interval of convergence, 

$\displaystyle -2< x< 2$

Therefore the radius of convergence is $\displaystyle 2$. 

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We can continue simplifying our most recent expression of $\displaystyle f(x)$. 

 $\displaystyle f(x)=\sum_{x=1}^{\infty}2x(-1)^n\left (\frac{x}{2} \right )^n$

 $\displaystyle f(x)=\sum_{n=1}^{\infty}(-1)^n\left(\frac{x^{n+1}}{2^{n-1}} \right )$

To test if this series diverges, before using a higher test, we may use the test for divergence.

The test for divergence informs that if the sequence does not approach 0 as n approaches infinity then the series diverges (NOTE: This only shows divergence, the converse is not true, that is, the test for divergence cannot be used to show convergence.).

We note that as  

$\displaystyle \lim_{x\rightarrow \infty }\frac{2x^2-x}{3x^2-7}=\frac{2}{3}$,

this is derived from the fact that to find the limit as x approaches infinity of a function, one must first find the horizontal asymptote. Since this function is a rational expression with the highest power in both the numerator and denominator, the horizontal asymptote is equal to the quotient of the leading coefficients of both the numerator and denominator, which in this case is 2/3.

Since the limit as x tends to infinity of this series is a nonzero value, we may conclude that the series diverges by the Test for Divergence.  

$\displaystyle \sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ converges due to the comparison test.

We start with the equation $\displaystyle \frac{1}{k^2} =\frac{1}{k^2}$. Since $\displaystyle sin(k) \leq 1$ for all values of k, we can multiply both side of the equation by the inequality and get $\displaystyle \frac{sin(k)}{k^2} \leq \frac{1}{k^2}$ for all values of k. Since $\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^2}$ is a convergent p-series with $\displaystyle p=2$, $\displaystyle \sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$  hence also converges by the comparison test.

We will use the Limit Comparison Test to show this result.

We first denote the genera term of the series by:

$\displaystyle u_{n}=\frac{9}{(9n+3)(9n+89)}$ and $\displaystyle v_{n}=\frac{9}{n^2}$.

We have $\displaystyle \lim_{n\rightarrow\infty}\frac{u_n}{v_{n}}=1$ and the series have the same nature .

We know that

$\displaystyle \sum_{n=1}^{\infty} u_{n}=\sum_{n=1}^{\infty} \frac{9}{n^2}$   is convergent by comparing the integral

$\displaystyle 9\int_{1}^{\infty} \frac{1}{x^2}dx$ which we know is convergent.

Therefore by the Limit Comparison Test.

we have $\displaystyle \sum_{n=1}^{\infty} v_{n}=\sum_{n=1}^{\infty} \frac{9}{(9n+3)(9n+89)}$.

If the series converges, then we know the terms must approach zero. At some point, the terms will be less than 1, meaning when you take the third power of the term, it will be less than the original term.

Other answers are not true for a convergent series by the $\displaystyle n^{th}$ term test for divergence.

In addition, the limit of the $\displaystyle n^{th}$ partial sums refers to the value the series converges to. A convergent series need not converge to zero. The alternating harmonic series is a good counter example to this.