which of the following statement are true? 
My attempts ; i take f(n) = sin(n)/n  if  i  take this in the form 
f(t) = sint/t, then it will satisfied  the all   above condition  that is 
sequence will be bounded and convergent  and series will also be convergent
 because 
sinx/x= (1/x)(x- x^3/3! +x^5/5! +......is convergent 
from my point of view all option a) ,option b) and option c) all are correct,,
as far  im very weak in real analysis .whethear my answer is correct or not,,,im very confused and pliz tell me the solution
 A: Hint:

For $(a)$, construct a function (first draw it) which spikes upwards to height $n$ at each positive integer value of $n$, but for which the spikes get thinner and thinner so that the sum of the areas of the spikes converges.

The same example will kill $(b)$ and $(c)$.
A: To give an example of what the other answer is talking about, consider the piece-wise function that is $0$ except near the integers where it's graph raises up to form triangles of height $n$. If the length base of the triangle at $n$ is given by $\frac{1}{n^3}$, then the area of each triangle is $\frac{1}{2n^2}$ and so the integral is $\frac{\pi^2}{12}$ by a well-known theorem.
However, $f(n)=n$ so none of the possibilities hold.

If you're having trouble picturing this, let me give you drawing instructions


*

*Draw a horizontal line at $x=0$. For most of the real line, the value of the function will be $0$.

*Put a dot at $(n,n)$ for every $n\in\mathbb{N}$. We want our function to satisfy $f(n)=n$ so that it violates a-c

*Put a dot at $(0.5, 0)$ and $(1.5,0)$ and erase the line between them. Instead, connect those two points to the point at $(1,1)$ to form a triangle shape (missing the bottom).

*Do the same with the pair of points $(2-\frac{1}{16}, 0)$ and $(2+\frac{1}{16},0)$.

*Do the same with the pair of points $(3-\frac{1}{54}, 0)$ and $(3+\frac{1}{54},0)$. 

*For every $n$, do the same with $(n-\frac{1}{2n^3},0)$ and $(n+\frac{1}{2n^3},0)$


You'll probably struggle to actually draw this due to precision issues, but if you get fine grid paper you can likely draw two of the triangles. Hopefully the construction that you would follow is clear.
