Upload your solutions to gradescope. For this assignment, make sure to show all relevant work. No points are given for correct solutions that have no justification or completely incorrect justification. You may be able to guess correctly if a series converges, but you will only receive points if your work justifies that conclusion.To demonstrate the expected level of work, suppose I had the following problem:limn4n35n37n3You may be able to see that the answer is 15 just by looking at the highest order terms. However, in this class, you need an extra step to fully justify. To get full points, your solution would look something like this:limn4n35n37n3=limn4n35n37n31n31n3=limn451n4n=4500=45If you want to solve a problen about convergence, you must specify which test you are using and inclade all relevant details in your solution. It in best to include complete sentences when necosary. For exanple, if you are anked to show whether the seriesn=11n21is comvergent or divergent, one possible solution is the following: We wse the integral test. Consider the fuactionf(x)=11x2First, note that f is continnons and poxitive on [1,). Furtheranore, f'(x)=-2x(1x2)-2=-2x(1x2)2, and for x>1, this is always negative, so f is decreasing. Hence, we may use the integral test. Now, we compute111x2dx=limt1111x2dx=limt(arctan(t)-arctan(1))=2-4Thus, the integral converges, so since an=1n21=f(n), we nse the integral test to conclude thatn=11n21comerges as well.In the following. you may use the fact that -1n converges if p>1 and diverges otherwise (you don't need to poone this if you use it, unless the problem were asking you to do that.) You can also we the fact that geometric sums converge for |r|<1 and use the formula for sum without proving it.find following limit, or show that it diverges:limxnarctan(13n)determine if folloning converges of diverger. converges, compute sum.n =54(-13)n-3Consider seriesn =31n(n2)Do following:(a) find a nth partial suarsn=i=3n1i(i2)(b) limitlimnsn =n=31n(n2)For series, determine they are absolutely convergent, conditionally divergent.(a)n =1(-1)nn2n3(b)n=1n3-3n2n53n24Find an approximation each series such error is no greater than 0.05(calculators computers encouraged this problem, but make sure to write down what your calculations how you got them). full credit, must both get correct answer demonstrate whyhas less 0.05.(a)n =11n3(b)n=1(-1)n-1n2