Return Razor partial View as JsonResult, ASP.NET MVC 5

Question

I am using jQuery datatables. This tutorial http://www.codeproject.com/Articles/165410/ASP-NET-MVC-Editable-DataTable-jQuery-DataTables-a shows how to call a controller action, using datatable. But, datatable expects a json response and I have a partial view to load the table rows. How to convert a ASP.NET MVC5 Razor (Partial)View(.cshtml) to JSON? Ideally using Newtonsoft.Json but other libraries will work.

I could load html table using json as shown in the tutorial

$(document).ready(function () {   
            $("#example").html();
            $("#example").DataTable({
                "processing": true, // for show progress bar
                "serverSide": true, // for process server side
                "filter": false, // this is for disable filter (search box)
                "orderMulti": false, // for disable multiple column at once
                "ajax": {
                    "url": "/home5/AjaxGetJsonData",
                    "type": "GET",
                    //"datatype": "html"
                    "datatype": "json"
                }
                ,
                "columns": [
                         { "data": "Name", "name": "ContactName", "autoWidth": true },
                        { "data": "Age", "name": "CompanyName", "autoWidth": true },
                        { "data": "DoB", "name": "Phone", "autoWidth": true }
                ]
            });
        });

With the controller returning

   return Json(dataTableData, JsonRequestBehavior.AllowGet);

I want to do something like

PartialViewResult searchView = PartialView("Search", model);
return Json(JsonConvert.SerializeObject(searchView)); 

I also tried converting the partial (.cshtml) view to string using below as the accepted answer here Render a view as a string

public string RenderRazorViewToString(string viewName, object model)
{
  ViewData.Model = model;
  using (var sw = new StringWriter())
  {
    var viewResult = ViewEngines.Engines.FindPartialView(ControllerContext,                                                                  viewName);
    var viewContext = new ViewContext(ControllerContext, viewResult.View,
                                 ViewData, TempData, sw);
    viewResult.View.Render(viewContext, sw);
    viewResult.ViewEngine.ReleaseView(ControllerContext, viewResult.View);
    return sw.GetStringBuilder().ToString();
  }
}