Recall the solution to one of your midterm exam problems

\(1\/2 + ∑\+\(n = 0\)\%∞\( 2 \((\(-1\))\)\^\(n + 1\)\)\/\(\(\ \)\(\((2  n + 1)\) \(π\)\(\ \ \)\)\)\ Cos \((\(\((2  n + 1)\) π\)\/2\ t)\)\)

To plot m terms of this in mathematica we might define

g[x_, m_] = 
        (1/2) + (∑_(n = 0)^m)(( 2 (((-1))^(n + 1)))/( (2  n + 1) π  )) Cos[(((2  n + 1) π)/2) t];

The semicolon surpresses Mathematica's output.  Plotting is then very easy.

Timing[Plot[{g[x, 1], g[x, 5]}, {t, 0, 8}, AspectRatio -> 1/3, 
          PlotStyle -> {RGBColor[1, 0, 0], 
              RGBColor[0, 0, 1]}];]

Defining g as above causes Mathematica to algebraicly manipulate the series into known functions (which in this case is an exact summation formula in terms of confluent hypergeometric functions).  During plotting these special functions must be evaluated, using up too much cpu time. In many cases it is much faster (but more cluttered) to just insert the series directly into the Plot function.

Timing[Plot[{(1/2) + (∑_(n = 0)^1)(( 2 (((-1))^(n + 1)))/( (2  n + 1) π  )) Cos[(((2  n + 1) π)/2) t], 
            (1/2) + (∑_(n = 0)^5)(( 2 (((-1))^(n + 1)))/( (2  n + 1) π  )) Cos[(((2  n + 1) π)/2) t]}, {t, 0, 8}, AspectRatio -> 1/3, 
          PlotStyle -> {RGBColor[1, 0, 0], 
              RGBColor[0, 0, 1]}];]

One nice convenience of using Mathematica is that most of the time it is able to choose the right number of grid points on which to plot.  For example it has no problem automatically resolving this highly oscillatory function

Plot[(∑_(n = 0)^8) Cos[(((2  n + 1) π)/2) t], {t, 
          0, 8}];

If you want to take control you can ask for more or less resolution using the PlotPoints option

Plot[(∑_(n = 0)^8) Cos[(((2  n + 1) π)/2) t], {t, 
          0, 8}, PlotPoints -> 2];

Plot[(∑_(n = 0)^8) Cos[(((2  n + 1) π)/2) t], {t, 
          0, 8}, PlotPoints -> 200];

The options MaxBend and PlotDivision are also useful in controling the accuracy of your plots (see Mathematica's built in help file for usage).
Mathematica also makes its best attempt to choose an optimal range over which to plot.  If you feel that it is truncating your plots or not zooming in on the interesting parts, you can use the PlotRange option to take control.

Plot[(∑_(n = 0)^8) Cos[(((2  n + 1) π)/2) t], {t, 
          0, 8}, PlotRange -> {-4, 10}];